Free math worksheets

- What derivative relationship defines an antiderivative? - Use the power rule for differentiation. - What happens to an additive constant when you differentiate? - Match coefficients and exponents after differentiating \(F\).

1. Because \(F\) is an antiderivative of \(f\), \(F'(x)=f(x)\). 2. For part a, \(F'(x)=kx^{k-1}\). Matching \(7x^6\) gives \(k=7\). 3. For part b, \(F'(x)=8x^3\). Matching \(kx^3\) gives \(k=8\). 4. For part c, \(F'(x)=\cos(x)\) for every real \(k\), because \(k^2\) is constant. Thus, any real value of \(k\) works.

a) \(k=7\) b) \(k=8\) c) Any \(k\in\mathbb{R}\), such as \(k=0\)

- What derivative relationship defines an antiderivative? - Which differentiation rule applies to a function inside another function? - Account for the derivative of the inner expression. - Differentiate before comparing the functions.

1. For part a, differentiate using the chain rule: \(F'(x)=\frac{1}{x^2+5}(2x)=\frac{2x}{x^2+5}=f(x)\). Therefore, \(F\) is an antiderivative of \(f\). 2. For part b, \(F'(x)=\cos(x^2)(2x)=2x\cos(x^2)=f(x)\). Therefore, \(F\) is an antiderivative of \(f\).

a) Yes, \(F\) is an antiderivative of \(f\). b) Yes, \(F\) is an antiderivative of \(f\).

- What derivative relationship defines an antiderivative? - Differentiate \(F\). - Which rule applies to a square root of a polynomial?

1. Rewrite \(F(x)=(2x^2+3)^{\frac{1}{2}}\). 2. Differentiate using the chain rule: \(F'(x)=\frac{1}{2}(2x^2+3)^{-\frac{1}{2}}(4x)\). 3. Simplify: \(F'(x)=\frac{2x}{\sqrt{2x^2+3}}=f(x)\). 4. Therefore, \(F\) is an antiderivative of \(f\).

Yes. Since \(F^{\prime}(x)=\frac{2x}{\sqrt{2x^2+3}}=f(x)\), \(F\) is an antiderivative of \(f\).

- Recall the derivative relationship between a function and its antiderivative. - Which rule is needed to differentiate a product? - Compare the entire derivative with \(g\).

1. Differentiate \(G(x)=x\sin(x)\) using the product rule. 2. \(G'(x)=\sin(x)+x\cos(x)\). 3. Since \(\sin(x)+x\cos(x)\ne\cos(x)\), \(G\) is not an antiderivative of \(g\).

No. \(G^{\prime}(x)=\sin(x)+x\cos(x)\ne g(x)\).

- Integrate each term separately. - Recall an antiderivative of \(\cos(x)\). - Different constants of integration produce different antiderivatives.

1. A general antiderivative is \(F(x)=6\sin(x)-2x^2+C\). 2. Choose two different constants, such as \(C=0\) and \(C=5\). This gives \(F_1(x)=6\sin(x)-2x^2\) and \(F_2(x)=6\sin(x)-2x^2+5\).

\(F_1(x)=6\sin(x)-2x^2\) and \(F_2(x)=6\sin(x)-2x^2+5\)

- What derivative relationship defines an antiderivative? - Apply the chain rule to the linear inner expressions. - Track the sign when differentiating cosine. - Constant terms disappear when differentiated.

1. For part a, differentiate using the chain rule: \(F'(x)=4e^{\frac{1}{2}x+1}\left(\frac{1}{2}\right)=2e^{\frac{1}{2}x+1}=f(x)\). Thus, \(F\) is an antiderivative of \(f\). 2. For part b, \(F'(x)=\frac{1}{2}[-\sin(2x)](2)=-\sin(2x)\). Since this is not \(f(x)=\sin(2x)\), \(F\) is not an antiderivative of \(f\).

a) Yes, \(F\) is an antiderivative of \(f\). b) No, \(F\) is not an antiderivative of \(f\).

- What equation must hold for \(F\) to be an antiderivative of \(f\)? - Use the chain rule. - Identify the inner and outer functions. - What is the derivative of cosine?

1. Differentiate \(F\) using the chain rule. 2. The derivative of the outer function \(2\cos(u)\) is \(-2\sin(u)\), and the derivative of the inner function \(u=x^3\) is \(3x^2\). 3. Thus, \(F'(x)=-2\sin(x^3)(3x^2)=-6x^2\sin(x^3)=f(x)\). 4. Therefore, \(F\) is an antiderivative of \(f\).

\(F^{\prime}(x)=-6x^2\sin(x^3)=f(x)\), so \(F\) is an antiderivative of \(f\).

- What equation must hold for \(F\) to be an antiderivative of \(f\)? - Split the fraction into separate powers before differentiating. - Apply the power rule to negative exponents.

1. Simplify \(F\): \(F(x)=2x^2-3x^{-2}\). 2. Differentiate using the power rule: \(F'(x)=4x+6x^{-3}\). 3. Since \(F'(x)=f(x)\), \(F\) is an antiderivative of \(f\).

Yes, \(F\) is an antiderivative of \(f\).

- Treat \(\frac1a\) as a constant factor. - Rewrite the variable in the denominator using a negative exponent. - Differentiate the given antiderivative and compare coefficients.

1. Rewrite \(h_a(x)=\frac1a x^{-2}\). Integrating on \(x>0\) gives \(H_a(x)=-\frac{1}{ax}+C\). 2. Differentiate the given function: \(H^{\prime}(x)=\frac{1}{5x^2}\). Comparing this with \(h_a(x)=\frac{1}{ax^2}\) gives \(a=5\).

1. \(H_a(x)=-\frac{1}{ax}+C\) 2. \(a=5\)

- Rewrite quotients using negative exponents. - Apply the power rule carefully to the exponent and coefficient. - Differentiate each proposed antiderivative to verify the match.

1. \(\int12(x+2)^{-4}\,\text{d}x=-4(x+2)^{-3}\), so A matches 1. 2. \(\int12x^{-4}\,\text{d}x=-4x^{-3}\), so B matches 2. 3. \(\int-12x^{-5}\,\text{d}x=3x^{-4}\), so C matches 3. 4. \(\int12(x+2)^{-5}\,\text{d}x=-3(x+2)^{-4}\), so D matches 4.

A–1; B–2; C–3; D–4

- Rewrite radicals and quotients using rational exponents. - Increase the exponent by \(1\), then divide by the new exponent. - Differentiate each result to check it.

1. Rewrite \(f^{\prime}(x)=\frac32x^{1/2}\). By the power rule, \(f(x)=x^{3/2}\). 2. Rewrite \(g^{\prime}(x)=\frac23x^{-1/3}\). By the power rule, \(g(x)=x^{2/3}\). 3. Rewrite \(h^{\prime}(x)=x^{-3/2}\). By the power rule, \(h(x)=-2x^{-1/2}=-\frac{2}{\sqrt{x}}\).

a) \(f(x)=x^{3/2}\) b) \(g(x)=x^{2/3}\) c) \(h(x)=-\frac{2}{\sqrt{x}}\)

- Rewrite radical denominators using negative exponents. - Integrate each term separately. - A constant derivative of \(-1\) comes from the term \(-x\).

1. By the power rule, \(f(x)=\frac{4/5}{2/5}x^{2/5}=2x^{2/5}\). 2. Rewrite the first term as \(\frac34x^{-3/4}\). Integrating term by term gives \(g(x)=3x^{1/4}-x\).

a) \(f(x)=2x^{2/5}\) b) \(g(x)=3\sqrt[4]{x}-x\)

- What derivative relationship defines an antiderivative? - Differentiate the sum term by term. - What is the derivative of \(\ln|x|\)? - Keep the constant factors when differentiating.

1. Differentiate term by term. 2. \(\frac{d}{dx}\left(\frac{1}{3}x^3\right)=x^2\). 3. For \(x\ne0\), \(\frac{d}{dx}[2\ln|x|]=\frac{2}{x}\). 4. Thus, \(F'(x)=x^2+\frac{2}{x}=f(x)\), so \(F\) is an antiderivative of \(f\).

\(F^{\prime}(x)=x^2+\frac{2}{x}=f(x)\), so \(F\) is an antiderivative of \(f\).

- First verify that the expression inside the logarithm is positive. - Differentiate the inner expression separately. - Apply \(\frac{d}{dx}\ln(u)=\frac{u'}{u}\).

1. Since \(x^2\ge0\) and \(e^x>0\), the expression \(x^2+e^x\) is positive for every real \(x\), so \(F\) is defined on all of \(\mathbb{R}\). 2. Let \(u(x)=x^2+e^x\). Then \(u'(x)=2x+e^x\). 3. By the chain rule, \(F'(x)=\frac{u'(x)}{u(x)}=\frac{2x+e^x}{x^2+e^x}=f(x)\). 4. Therefore, \(F\) is an antiderivative of \(f\) for every real \(x\).

Yes. Since \(x^2+e^x>0\) and \(F'(x)=\frac{2x+e^x}{x^2+e^x}=f(x)\), \(F\) is an antiderivative on \(\mathbb{R}\).

- Recall an antiderivative of \(\frac1x\) on \(x>0\). - Integrate each term separately. - Different constants produce different antiderivatives.

1. A general antiderivative is \(F(x)=6\ln(x)+2x+C\). 2. Choose two different constants, such as \(C=0\) and \(C=5\). Then \(F_1(x)=6\ln(x)+2x\) and \(F_2(x)=6\ln(x)+2x+5\).

\(F_1(x)=6\ln(x)+2x\) and \(F_2(x)=6\ln(x)+2x+5\)

- Reverse the power rule by increasing the exponent by \(1\) and dividing by the new exponent. - Choose each coefficient so that differentiating reproduces the given coefficient. - Differentiate each result to check it.

1. For a), reverse the power rule: \(f(x)=4x^3\). Check: \(f^{\prime}(x)=12x^2\). 2. For b), \(f(x)=\frac25x^5\). Check: \(f^{\prime}(x)=2x^4\). 3. For c), \(f(x)=\frac15x^4\). Check: \(f^{\prime}(x)=\frac45x^3\). 4. For d), \(f(x)=10x\). Check: \(f^{\prime}(x)=10\).

a) \(f(x)=4x^3\) b) \(f(x)=\frac25x^5\) c) \(f(x)=\frac15x^4\) d) \(f(x)=10x\)

- Recall the derivatives of sine and cosine. - Rewrite expressions with \(x\) in the denominator using negative exponents. - Integrate sums term by term. - Check the sign when working with sine and cosine.

1. Since the derivative of \(\sin(x)\) is \(\cos(x)\), one choice for a) is \(f(x)=5\sin(x)\). 2. Since the derivative of \(\cos(x)\) is \(-\sin(x)\), one choice for b) is \(f(x)=\cos(x)\). 3. Rewrite \(\frac{2}{x^2}\) as \(2x^{-2}\). Reversing the power rule gives \(f(x)=-2x^{-1}=-\frac2x\). 4. Integrating term by term gives \(f(x)=\frac13x^3-\frac12x^2\).

a) \(f(x)=5\sin(x)\) b) \(f(x)=\cos(x)\) c) \(f(x)=-\frac2x\) d) \(f(x)=\frac13x^3-\frac12x^2\)

- Apply the reverse power rule to each term. - A constant derivative comes from a linear term. - Differentiate your result to check it.

1. Integrate each term using the reverse power rule: \(12x^3\) gives \(3x^4\), \(-5x^4\) gives \(-x^5\), and \(2\) gives \(2x\). 2. Thus one possible function is \(f(x)=3x^4-x^5+2x\). 3. Differentiating gives \(f^{\prime}(x)=12x^3-5x^4+2\), as required.

\(f(x)=3x^4-x^5+2x\). More generally, \(f(x)=3x^4-x^5+2x+C\), where \(C\in\mathbb{R}\).

- Integrate the two terms separately. - Which trigonometric function has derivative \(\sin(x)\)? Pay attention to the sign. - Reverse the power rule for \(4x^7\). - Differentiate your result to verify it.

1. Since the derivative of \(-8\cos(x)\) is \(8\sin(x)\), the first term has antiderivative \(-8\cos(x)\). 2. Reversing the power rule gives an antiderivative of \(4x^7\) as \(\frac12x^8\). 3. Therefore, one possible function is \(g(x)=-8\cos(x)+\frac12x^8\). 4. Differentiating gives \(g^{\prime}(x)=8\sin(x)+4x^7\), as required.

\(g(x)=-8\cos(x)+\frac12x^8\). More generally, \(g(x)=-8\cos(x)+\frac12x^8+C\), where \(C\in\mathbb{R}\).

- Rewrite the fraction using a negative exponent. - Recall which trigonometric function has derivative \(\cos(x)\). - Reverse the power rule, then differentiate your result to check it.

1. Rewrite \(\frac{4}{x^5}\) as \(4x^{-5}\). 2. An antiderivative of \(\cos(x)\) is \(\sin(x)\). 3. Reversing the power rule gives an antiderivative of \(4x^{-5}\) as \(-x^{-4}=-\frac1{x^4}\). 4. Therefore, one possible function is \(g(x)=\sin(x)-\frac1{x^4}\).

\(g(x)=\sin(x)-\frac1{x^4}\) for \(x\ne0\)

- Use the power rule in reverse to find the general antiderivative. - A point on the graph gives an equation involving \(C\). - Substitute each point separately.

1. The general antiderivative is \(F(x)=2x^3-2x^2+x+C\). 2. For part a, \(F(0)=7\) gives \(C=7\). 3. For part b, \(F(1)=2\) gives \(1+C=2\), so \(C=1\). 4. For part c, \(F(-1)=0\) gives \(-2-2-1+C=0\), so \(C=5\).

a) \(F(x)=2x^3-2x^2+x+7\) b) \(F(x)=2x^3-2x^2+x+1\) c) \(F(x)=2x^3-2x^2+x+5\)

- Rewrite terms with \(x\) in the denominator using negative exponents. - Split a quotient with a sum in the numerator into separate terms. - Apply the reverse power rule term by term.

1. For a), rewrite the function as \(x^4-2x^{-3}+1\). Integrating term by term gives \(F(x)=\frac15x^5+x^{-2}+x=\frac15x^5+\frac1{x^2}+x\). 2. For b), simplify first: \(\frac{x^3+5}{x^2}=x+5x^{-2}\). Integrating term by term gives \(F(x)=\frac12x^2-5x^{-1}=\frac12x^2-\frac5x\).

a) \(F(x)=\frac15x^5+\frac1{x^2}+x\) b) \(F(x)=\frac12x^2-\frac5x\)

- Rewrite radicals using rational exponents. - Simplify the quotient in b) before integrating. - Apply the reverse power rule carefully to fractional exponents.

1. For a), rewrite the function as \(3x^{1/2}-x^{-1/2}\). Integrating term by term gives \(F(x)=2x^{3/2}-2x^{1/2}=2x^{3/2}-2\sqrt{x}\). 2. For b), simplify first: \(\frac{\sqrt{x}+2}{\sqrt{x}}=1+2x^{-1/2}\). Integrating gives \(F(x)=x+4x^{1/2}=x+4\sqrt{x}\).

a) \(F(x)=2x^{3/2}-2\sqrt{x}\) b) \(F(x)=x+4\sqrt{x}\)

- First find the general antiderivative. - All antiderivatives of the same function differ by a constant. - Choose three different values for that constant.

1. Integrating term by term gives the general antiderivative \(F(x)=2x^4-2x^3+5x+C\). 2. Choose three different constants, such as \(C=0\), \(C=3\), and \(C=-1\). 3. This gives \(F_1(x)=2x^4-2x^3+5x\), \(F_2(x)=2x^4-2x^3+5x+3\), and \(F_3(x)=2x^4-2x^3+5x-1\).

\(F_1(x)=2x^4-2x^3+5x\) \(F_2(x)=2x^4-2x^3+5x+3\) \(F_3(x)=2x^4-2x^3+5x-1\)

- Expand the squared binomial before integrating. - Integrate each term of the resulting polynomial separately. - Use the reverse power rule.

1. Expand the square: \(f(x)=\frac12x^4-3x^2+\frac92\). 2. Integrate term by term to obtain \(F(x)=\frac1{10}x^5-x^3+\frac92x\). 3. Differentiating this expression reproduces \(f(x)\).

\(F(x)=\frac1{10}x^5-x^3+\frac92x\)

- Recall the antiderivatives of sine and cosine. - Rewrite reciprocal powers and radicals using exponents. - Integrate each term separately and keep constant factors.

1. For a), integrate term by term: \(F(x)=-5\cos(x)+\frac15x^5\). 2. For b), rewrite \(\frac3{x^2}\) as \(3x^{-2}\). Integrating gives \(F(x)=-\frac3x-2\sin(x)\). 3. For c), integrate inside the parentheses and retain the factor \(4\): \(F(x)=4\sin(x)+8\sqrt{x}\).

a) \(F(x)=-5\cos(x)+\frac15x^5\) b) \(F(x)=-\frac3x-2\sin(x)\) c) \(F(x)=4\sin(x)+8\sqrt{x}\)

- Differentiate the proposed antiderivative. - What must the derivative of an antiderivative equal? - What effect does an additive constant have on a derivative?

1. Differentiate \(F\): \(F'(x)=4x^3-6x=f(x)\). Thus, \(F\) is an antiderivative of \(f\). 2. A function is an antiderivative of \(f\) on an interval when its derivative equals \(f\) at every point of the interval. 3. Differentiate \(H\): \(H'(x)=4x^3-6x=f(x)\). The different constant term has no effect on the derivative. 4. Therefore, \(H\) is also an antiderivative of \(f\).

a) Yes, because \(F'(x)=4x^3-6x=f(x)\). b) An antiderivative has derivative \(f\). Since \(H'(x)=4x^3-6x=f(x)\), \(H\) is also an antiderivative.

- For \(e^{ax+b}\), divide by the derivative of the exponent. - Integrate sums term by term. - Differentiate each result to check the inner-factor adjustment.

1. For a), account for the inner derivative \(2\): \(F(x)=2e^{2x}\). 2. For b), the inner derivative of \(5-x\) is \(-1\), so an antiderivative of \(e^{5-x}\) is \(-e^{5-x}\). Therefore, \(F(x)=-e^{5-x}+\frac13x^3\). 3. For c), account for the inner derivative \(4\): \(F(x)=\frac18e^{4x-2}-3x\).

a) \(F(x)=2e^{2x}\) b) \(F(x)=-e^{5-x}+\frac13x^3\) c) \(F(x)=\frac18e^{4x-2}-3x\)

- Use the power rule for antiderivatives. - What effect does an additive constant have on the derivative? - A point on the graph gives an equation for \(C\). - Substitute \(x=1\) and the given output.

1. Rewrite \(f(x)=x^{-2}+1\). The family of antiderivatives is \(F(x)=-\frac{1}{x}+x+C\). 2. Four examples are obtained by choosing different constants, such as \(C=0, 1, -5, 10\). 3. For part b, use \(F(1)=0\): \(-1+1+C=0\), so \(C=0\). 4. Therefore, the required antiderivative is \(F(x)=-\frac{1}{x}+x\).

a) Examples: \(F_1(x)=-\frac{1}{x}+x\), \(F_2(x)=-\frac{1}{x}+x+1\), \(F_3(x)=-\frac{1}{x}+x-5\), and \(F_4(x)=-\frac{1}{x}+x+10\) b) \(F(x)=-\frac{1}{x}+x\)

- Use the chain rule to differentiate the candidate antiderivatives. - Look for a numerator that is the derivative of the denominator. - For a linear inner function, account for its constant derivative. - Differentiate each candidate to check your matches.

1. Differentiate \(F_C\): \(F_C'(x)=4\cos(4x-2)=f_1(x)\). Thus \(f_1\) matches \(F_C\). 2. Since the derivative of \(x^2+x\) is \(2x+1\), \(\frac{d}{dx}\ln|x^2+x|=\frac{2x+1}{x^2+x}=f_2(x)\). Thus \(f_2\) matches \(F_A\). 3. Differentiate \(F_D\): \(F_D'(x)=\frac{1}{10}\cdot 5(2x-3)^4\cdot 2=(2x-3)^4=f_3(x)\). Thus \(f_3\) matches \(F_D\). 4. Differentiate \(F_B\): \(F_B'(x)=2\cdot 0.5e^{0.5x}=e^{0.5x}=f_4(x)\). Thus \(f_4\) matches \(F_B\).

1. C 2. A 3. D 4. B

- Identify the function variable in each part. - Treat all other symbols as constants. - Rewrite a variable in the denominator using a negative exponent before applying the power rule. - Differentiate your result to check it.

1. Integrate with respect to \(x\): \(F(x)=\frac{4}{k}\cdot\frac{x^4}{4}-3a\cdot\frac{x^3}{3}=\frac{x^4}{k}-ax^3\). 2. Rewrite \(\frac{k}{t^2}=kt^{-2}\) and integrate with respect to \(t\): \(G(t)=-kt^{-1}+at=-\frac{k}{t}+at\). 3. Integrate with respect to \(a\), treating \(x\) as a constant: \(H(a)=\frac1x\cdot\frac{a^2}{2}+a\sqrt{x}=\frac{a^2}{2x}+a\sqrt{x}\).

a) \(F(x)=\frac{x^4}{k}-ax^3\) b) \(G(t)=-\frac{k}{t}+at\) c) \(H(a)=\frac{a^2}{2x}+a\sqrt{x}\)

- Check the sign when integrating \(-\sin(x)\). - The exponential function \(e^t\) is its own antiderivative. - Apply the power rule to fractional exponents. - Differentiate each result to verify it.

1. Integrate with respect to \(x\): \(F(x)=\frac1k\cos(x)+k^2x\). 2. Integrate with respect to \(t\): \(G(t)=e^t-\frac12k^2t^2\). 3. Apply the power rule with respect to \(z\): \(H(z)=\frac25z^{5/2}+\frac52a^2z\).

a) \(F(x)=\frac1k\cos(x)+k^2x\) b) \(G(t)=e^t-\frac12k^2t^2\) c) \(H(z)=\frac25z^{5/2}+\frac52a^2z\)

- Which integration rules apply to the two terms? - What equation follows from the point on the graph? - How can you use that equation to find \(C\)? - Keep the logarithm's domain in mind.

1. Integrate each term: \(F(x)=\frac{1}{2}e^{2x}+\ln(x+2)+C\). The domain condition \(x>-2\) makes \(x+2>0\). 2. Use \(F(0)=1.5\): \(\frac{1}{2}+\ln(2)+C=1.5\). 3. Solve for the constant: \(C=1-\ln(2)\). 4. Therefore, \(F(x)=\frac{1}{2}e^{2x}+\ln(x+2)+1-\ln(2)\).

\(F(x)=\frac{1}{2}e^{2x}+\ln(x+2)+1-\ln(2)\)

- First find the general antiderivative of \(g\). - Determine the range of the cosine term. - What must be true of the minimum and maximum for the function to have both signs? - Solve the resulting inequalities for \(C\).

1. The general antiderivative is \(G(x)=-2\cos(3x+2)+C\). 2. Because cosine ranges from \(-1\) to \(1\), the term \(-2\cos(3x+2)\) ranges from \(-2\) to \(2\). Thus the range of \(G\) is \([C-2, C+2]\). 3. For \(G\) to take both positive and negative values, its minimum must be below \(0\) and its maximum above \(0\): \(C-2<0\) and \(C+2>0\). 4. These inequalities give \(-2<C<2\).

\(G(x)=-2\cos(3x+2)+C\), where \(C\in(-2, 2)\)

- Find an antiderivative of each term separately. - Check the sign that comes from the derivative of \(2-x\). - Use the given value to solve for the integration constant. - Differentiate your result to verify it.

1. Integrate each term: \(F(x)=\frac{1}{8}x^4+e^{2-x}+C\). 2. Apply \(F(2)=10\): \(\frac{1}{8}(2)^4+e^0+C=10\). 3. Simplify: \(2+1+C=10\), so \(C=7\). 4. Therefore, \(F(x)=\frac{1}{8}x^4+e^{2-x}+7\).

\(F(x)=\frac{1}{8}x^4+e^{2-x}+7\)

- Which function has a derivative of the form \(\frac{1}{x+a}\)? - Account for the linear inner function when integrating the cosine term. - Use \(F(0)=0\) to isolate \(C\). - A logarithm property can combine the two logarithmic terms.

1. Integrate each term: \(F(x)=2\ln(x+3)-2\sin\left(\frac{x}{2}\right)+C\). The domain condition ensures \(x+3>0\). 2. Apply \(F(0)=0\): \(2\ln(3)-2\sin(0)+C=0\). 3. Therefore, \(C=-2\ln(3)\). 4. The required antiderivative is \(F(x)=2\ln(x+3)-2\sin\left(\frac{x}{2}\right)-2\ln(3)\).

\(F(x)=2\ln(x+3)-2\sin\left(\frac{x}{2}\right)-2\ln(3)\), equivalently \(F(x)=2\ln\left(\frac{x+3}{3}\right)-2\sin\left(\frac{x}{2}\right)\)

- Integrate the exponential and polynomial terms separately. - How does the derivative of \(3x+3\) affect the exponential antiderivative? - Substitute the given input and output to find \(C\). - Solve the resulting equation for the unknown constant.

1. Integrate term by term: \(F(x)=\frac{2}{3}e^{3x+3}-2x^2+C\). 2. Apply \(F(-1)=5\): \(\frac{2}{3}e^0-2(-1)^2+C=5\). 3. Simplify: \(\frac{2}{3}-2+C=5\), so \(-\frac{4}{3}+C=5\). 4. Thus, \(C=\frac{19}{3}\), and \(F(x)=\frac{2}{3}e^{3x+3}-2x^2+\frac{19}{3}\).

\(F(x)=\frac{2}{3}e^{3x+3}-2x^2+\frac{19}{3}\)

- How do you differentiate a composite function? - What equation must hold for one function to be an antiderivative of another? - Compose the functions before integrating. - How does a specified function value determine the integration constant?

1. For part a, compose the functions: \(h(x)=f(g(x))=\left(\frac{1}{3}x+2\right)^2-5\). 2. Differentiate \(H\): \(H'(x)=3\left(\frac{1}{3}x+2\right)^2\left(\frac{1}{3}\right)-5=\left(\frac{1}{3}x+2\right)^2-5=h(x)\). Therefore, \(H\) is an antiderivative of \(h\). 3. For part b, \(k(x)=6g(f(x))=6\left(\frac{1}{3}(x^2-5)+2\right)=2x^2+2\). 4. The general antiderivative is \(K(x)=\frac{2}{3}x^3+2x+C\). 5. Use \(K(3)=20\): \(\frac{2}{3}(3)^3+2\cdot 3+C=20\), so \(24+C=20\) and \(C=-4\). 6. Thus, \(K(x)=\frac{2}{3}x^3+2x-4\).

a) \(H'(x)=\left(\frac{1}{3}x+2\right)^2-5=f(g(x))\), so \(H\) is an antiderivative of \(h\). b) \(K(x)=\frac{2}{3}x^3+2x-4\)

- Use the condition \(F'(x)=f(x)\). - Pay close attention to the chain rule in part c. - In part b, compare exponents first. - Also verify that the coefficients match.

1. In each part, require \(F'(x)=f(x)\). 2. For part a, \(F'(x)=3ax^2\). Matching coefficients with \(\frac{1}{3}x^2\) gives \(3a=\frac{1}{3}\), so \(a=\frac{1}{9}\). 3. For part b, \(F'(x)=4x^3\). Matching exponents gives \(a-2=3\), so \(a=5\). Then \(a-1=4\), so the coefficient also matches. 4. For part c, \(F'(x)=3ae^{3x}\). Matching coefficients with \(e^{3x}\) gives \(3a=1\), so \(a=\frac{1}{3}\).

a) \(a=\frac{1}{9}\) b) \(a=5\) c) \(a=\frac{1}{3}\)

- Which differentiation rule applies to the product in part a? - Recall the product rule. - Rewrite the fraction in part b using a negative exponent. - Track the negative sign and the derivative of the inner expression.

1. For part a, use the product rule: \(F'(x)=e^x+(x-1)e^x=xe^x=f(x)\). The statement is true. 2. For part b, write \(F(x)=(x^2+1)^{-1}\). Then \(F'(x)=-(x^2+1)^{-2}(2x)=-\frac{2x}{(x^2+1)^2}\), which is not equal to \(f(x)\). The statement is false.

a) True, because \(F'(x)=xe^x=f(x)\). b) False, because \(F'(x)=-\frac{2x}{(x^2+1)^2}\ne f(x)\).

- Differentiate each proposed antiderivative. - Use the power rule in reverse for functions that have no match. - An antiderivative \(F\) must satisfy \(F'=f\). - You may choose any additive constant when none is specified.

1. Differentiate the listed antiderivatives: \(F_1'(x)=2x+3\), \(F_2'(x)=6x^2\), \(F_3'(x)=4x^3-x\), and \(F_4'(x)=5x^4\). 2. Therefore, the matches are A-2, B-1, D-3, and E-4. 3. For C, one antiderivative is \(\int \frac{1}{2}x^3\,\text{d}x=\frac{1}{8}x^4\). 4. For F, one antiderivative is \(\int 3\,\text{d}x=3x\).

Matches: A-2, B-1, D-3, E-4. For C: \(F(x)=\frac{1}{8}x^4\) For F: \(F(x)=3x\)

- How are derivatives and antiderivatives related? - First write each general antiderivative with a constant \(C\). - Substitute the coordinates of the given point to determine \(C\).

1. For part a, \(G_A'(x)=4x-1=g_1(x)\), and \(G_B'(x)=2x^3-\frac{1}{2}x^2=g_4(x)\). 2. For \(g_2(x)=3x^2-4x\), the general antiderivative is \(G_2(x)=x^3-2x^2+C\). Using \(G_2(1)=2\) gives \(-1+C=2\), so \(C=3\). Thus, \(G_2(x)=x^3-2x^2+3\). 3. For \(g_3(x)=x^3\), the general antiderivative is \(G_3(x)=\frac{1}{4}x^4+C\). Using \(G_3(1)=2\) gives \(\frac{1}{4}+C=2\), so \(C=\frac{7}{4}\). Thus, \(G_3(x)=\frac{1}{4}x^4+\frac{7}{4}\).

a) \(G_A\) matches \(g_1\), and \(G_B\) matches \(g_4\). b) \(G_2(x)=x^3-2x^2+3\); \(G_3(x)=\frac{1}{4}x^4+\frac{7}{4}\)

- How is the slope of \(F\) related to \(f\)? - When are two tangent lines parallel? - What equation gives all points with the same slope? - Use the given point to determine \(C\).

1. The slope of \(F\) at \(x=1\) is \(F'(1)=f(1)=1-4+5=2\). 2. Parallel tangent lines have equal slopes, so solve \(f(x)=2\): \(x^2-4x+5=2\). 3. This becomes \(x^2-4x+3=0\), or \((x-1)(x-3)=0\). Besides \(x=1\), the other solution is \(x=3\). 4. For part c, the general antiderivative is \(G(x)=\frac{1}{3}x^3-2x^2+5x+C\). 5. Apply \(G(3)=8\): \(9-18+15+C=8\), so \(6+C=8\) and \(C=2\). 6. Therefore, \(G(x)=\frac{1}{3}x^3-2x^2+5x+2\).

a) \(2\) b) Yes. The other point has x-coordinate \(x=3\). c) \(G(x)=\frac{1}{3}x^3-2x^2+5x+2\)

- Split the fraction into separate terms. - Check the sign when integrating \(\sin(x)\). - Use the given point to determine the constant of integration.

1. Rewrite \(g(x)=\frac35\sin(x)-\frac25\). A general antiderivative is \(G(x)=-\frac35\cos(x)-\frac25x+C\). 2. For part a, choose \(C=0\): \(G_1(x)=-\frac35\cos(x)-\frac25x\). 3. For part b, use \(G_2(0)=1\): \(-\frac35+C=1\), so \(C=\frac85\). Therefore, \(G_2(x)=-\frac35\cos(x)-\frac25x+\frac85\).

a) \(G_1(x)=-\frac35\cos(x)-\frac25x\) b) \(G_2(x)=-\frac35\cos(x)-\frac25x+\frac85\)

- What equation must hold for \(f\) to be an antiderivative of \(g\)? - Use the product rule on the nonconstant term. - What happens to the constant \(5\) when differentiated? - Factor out the common exponential term after differentiating.

1. Differentiate the product \((x^3-4x)e^x\) using the product rule. The constant \(5\) differentiates to \(0\). 2. \(f'(x)=(3x^2-4)e^x+(x^3-4x)e^x\). 3. Factor out \(e^x\): \(f'(x)=(x^3+3x^2-4x-4)e^x\). 4. This equals \(g(x)\), so \(f\) is an antiderivative of \(g\).

\(f^{\prime}(x)=(3x^2-4)e^x+(x^3-4x)e^x=(x^3+3x^2-4x-4)e^x=g(x)\). Therefore, \(f\) is an antiderivative of \(g\).

- What derivative relationship defines an antiderivative? - Apply the product rule separately to the two products. - Track the signs of the trigonometric derivatives carefully. - Look for terms that cancel after differentiation.

1. Differentiate both terms of \(f\) using the product rule. 2. The derivative of \((x^2-2)\cos(x)\) is \(2x\cos(x)-(x^2-2)\sin(x)\). 3. The derivative of \(-2x\sin(x)\) is \(-2\sin(x)-2x\cos(x)\). 4. Combine the results: \(f'(x)=2x\cos(x)-(x^2-2)\sin(x)-2\sin(x)-2x\cos(x)\). 5. The cosine terms cancel, and \(2\sin(x)-2\sin(x)=0\), leaving \(f'(x)=-x^2\sin(x)=g(x)\). 6. Therefore, \(f\) is an antiderivative of \(g\).

\(f^{\prime}(x)=2x\cos(x)-(x^2-2)\sin(x)-2\sin(x)-2x\cos(x)=-x^2\sin(x)=g(x)\). Therefore, \(f\) is an antiderivative of \(g\).

- Does the expression resemble the result of a familiar differentiation rule? - Look for two terms that could come from differentiating a product. - Find functions \(u\) and \(v\) whose product has derivative \(f\). - Differentiate your proposed antiderivative to check it.

1. Recognize the product-rule pattern \(u'v+uv'\). 2. Choose \(u(x)=x^3\) and \(v(x)=\sin(x)\). Then \(u'(x)=3x^2\) and \(v'(x)=\cos(x)\). 3. Therefore, \(\frac{d}{dx}[x^3\sin(x)]=3x^2\sin(x)+x^3\cos(x)=f(x)\). 4. One antiderivative is \(F(x)=x^3\sin(x)\).

\(F(x)=x^3\sin(x)\)

- Use the chain rule: derivative of the outer function times derivative of the inner function. - Rewrite a square root as a power when helpful. - Identify the inner expression in each part. - Compare the complete derivative with \(f\).

1. For part a, use the chain rule: \(F'(x)=5(x^2-4x)^4(2x-4)\). This equals \(f(x)\), so \(F\) is an antiderivative of \(f\). 2. For part b, write \(F(x)=(x^2+1)^{\frac{1}{2}}+4\). Then \(F'(x)=\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)=\frac{x}{\sqrt{x^2+1}}=f(x)\). Thus, \(F\) is an antiderivative of \(f\).

a) Yes, \(F\) is an antiderivative of \(f\). b) Yes, \(F\) is an antiderivative of \(f\).

- Track the signs when differentiating and integrating sine and cosine. - Treat \(\pi\) as a constant. - Apply the power rule term by term.

1. Differentiate term by term: \(f^{\prime}(x)=3x^3+\sin(x)\). 2. Integrate term by term: \(F(x)=\frac{3}{20}x^5-\sin(x)+\pi x\).

\(f^{\prime}(x)=3x^3+\sin(x)\) \(F(x)=\frac{3}{20}x^5-\sin(x)+\pi x\)

- Expand the squared binomial first. - Apply the derivative and antiderivative power rules term by term. - Differentiate the antiderivative to check it.

1. Expand the function: \(f(x)=x^2+4x+4-3x^3\). 2. Differentiate term by term: \(f^{\prime}(x)=-9x^2+2x+4\). 3. Integrate term by term: \(F(x)=-\frac34x^4+\frac13x^3+2x^2+4x\).

\(f^{\prime}(x)=-9x^2+2x+4\) \(F(x)=-\frac34x^4+\frac13x^3+2x^2+4x\)

- Differentiate and integrate each term separately. - Track the signs for sine and cosine. - Remember that \(e^x\) is its own derivative and antiderivative. - Use the first derivative to find slope at a point.

1. Differentiate term by term: \(g'(x)=-3\sin x-4e^x\). Differentiating again gives \(g''(x)=-3\cos x-4e^x\). 2. Integrate term by term. One antiderivative is \(G(x)=3\sin x-4e^x\). More generally, \(G(x)=3\sin x-4e^x+C\). 3. The slope at \(x=0\) is \(g'(0)=-3\sin 0-4e^0=-4\).

a) \(g'(x)=-3\sin x-4e^x\); \(g''(x)=-3\cos x-4e^x\) b) \(G(x)=3\sin x-4e^x\) c) The slope is \(-4\).

- Work term by term using the sum rule. - Use the basic derivative and antiderivative rules for powers, exponential functions, sine, and cosine. - Check the signs of the trigonometric terms. - Include an arbitrary constant in the general antiderivative.

1. Differentiate term by term: \(f'(x)=4e^x-2x^3+\cos x\). 2. Differentiate again: \(f''(x)=4e^x-6x^2-\sin x\). 3. Integrate the original function term by term: \(F(x)=4e^x-\frac{1}{10}x^5-\cos x+C\). Differentiating this expression returns \(f(x)\).

\(f'(x)=4e^x-2x^3+\cos x\) \(f''(x)=4e^x-6x^2-\sin x\) \(F(x)=4e^x-\frac{1}{10}x^5-\cos x+C\)

- Differentiate and integrate each term separately. - Include an arbitrary constant in the general antiderivative. - Substitute the coordinates of the given point into \(G\). - Use \(e^0=1\) to solve for the constant.

1. Differentiate term by term: \(g'(x)=4x-3e^x\) and \(g''(x)=4-3e^x\). 2. The general antiderivative is \(G(x)=\frac{2}{3}x^3-3e^x+x+C\). Use the condition \(G(0)=2\): \(-3+C=2\), so \(C=5\). Therefore, \(G(x)=\frac{2}{3}x^3-3e^x+x+5\).

a) \(g'(x)=4x-3e^x\); \(g''(x)=4-3e^x\) b) \(G(x)=\frac{2}{3}x^3-3e^x+x+5\)

- What derivative relationship defines an antiderivative? - Use the product rule. - Remember the chain-rule factor when differentiating \(e^{2x}\). - Factor out the common exponential term before comparing.

1. Use the product rule with \(u(x)=x^2-3x\) and \(v(x)=e^{2x}\). 2. Their derivatives are \(u'(x)=2x-3\) and \(v'(x)=2e^{2x}\). 3. Therefore, \(F'(x)=(2x-3)e^{2x}+2(x^2-3x)e^{2x}\). 4. Factor and simplify: \(F'(x)=e^{2x}(2x-3+2x^2-6x)=(2x^2-4x-3)e^{2x}=f(x)\). 5. Hence, \(F\) is an antiderivative of \(f\).

Yes. \(F^{\prime}(x)=(2x^2-4x-3)e^{2x}=f(x)\), so \(F\) is an antiderivative of \(f\).

- Use the zero-product property in part a. - Recall the quadratic formula. - Differentiate \(F\) to verify the antiderivative relationship. - Antiderivatives differ by a constant; use the initial value to find it.

1. For part a, \(e^x>0\), so the zeros satisfy \(x^2+4x+2=0\). 2. The quadratic formula gives \(x=\frac{-4\pm\sqrt{16-8}}{2}=-2\pm\sqrt{2}\). 3. For part b, differentiate using the product rule: \(F'(x)=(2x+2)e^x+(x^2+2x)e^x=(x^2+4x+2)e^x=f(x)\). 4. Every antiderivative has the form \(G(x)=F(x)+C\). Since \(F(-2)=0\), the condition \(G(-2)=1\) gives \(C=1\). 5. Therefore, \(G(x)=(x^2+2x)e^x+1\).

a) \(x=-2-\sqrt{2}\) and \(x=-2+\sqrt{2}\) b) \(F'(x)=f(x)\), and \(G(x)=(x^2+2x)e^x+1\)

- Differentiate each candidate and compare it with \(f\). - Use the product rule for a polynomial times \(e^x\). - What happens to an added constant when differentiated? - Compare the polynomial coefficients after factoring out \(e^x\).

1. Differentiate each product using the product rule. 2. \(F_A'(x)=(x^2+7x+10)e^x\ne f(x)\). 3. \(F_B'(x)=(x^2+6x+5)e^x\ne f(x)\). 4. \(F_C'(x)=(x^2+5x+5)e^x=f(x)\). The constant \(4\) differentiates to \(0\). 5. \(F_D'(x)=(x^2+9x+17)e^x\ne f(x)\). 6. Therefore, \(F_C\) is an antiderivative of \(f\).

\(F_C(x)=(x^2+3x+2)e^x+4\)

- Use \(F'(x)=f(x)\). - Differentiate with the product rule and factor out \(e^x\). - Equal polynomials have matching coefficients. - Solve the resulting three equations in order.

1. Differentiate \(F\) using the product rule: \(F'(x)=(2ax+b)e^x+(ax^2+bx+c)e^x\). 2. Combine terms: \(F'(x)=[ax^2+(2a+b)x+(b+c)]e^x\). 3. Match coefficients with \((x^2-x-2)e^x\): \(a=1\), \(2a+b=-1\), and \(b+c=-2\). 4. From \(a=1\), \(2+b=-1\), so \(b=-3\). Then \(-3+c=-2\), so \(c=1\). 5. Therefore, \(a=1\), \(b=-3\), and \(c=1\).

\(a=1\), \(b=-3\), \(c=1\)

- Which integration rules apply to the exponential, polynomial, and sine terms? - How does the coefficient in the exponent affect the antiderivative? - What equation does the point on the graph provide? - Use that equation to determine \(C\).

1. Integrate term by term: \(F(x)=2e^{2x}-\frac{1}{6}x^3-\cos(x)+C\). 2. Use \(F(0)=3\): \(2e^0-0-\cos(0)+C=3\). 3. This gives \(2-1+C=3\), so \(C=2\). 4. Therefore, \(F(x)=2e^{2x}-\frac{1}{6}x^3-\cos(x)+2\).

\(F(x)=2e^{2x}-\frac{1}{6}x^3-\cos(x)+2\)

- Differentiate \(F\) and compare the result with \(f\). - Expand and simplify \(F\) before differentiating. - Track signs carefully when differentiating negative powers. - Compare the numerators after using a common denominator.

1. Expand and simplify: \(F(x)=\frac{x^4-8x^2+16}{x}=x^3-8x+16x^{-1}\). 2. Differentiate: \(F'(x)=3x^2-8-16x^{-2}\). 3. With a common denominator, \(F'(x)=\frac{3x^4-8x^2-16}{x^2}\). 4. The constant term in the numerator is \(-16\), not \(+16\). Therefore, \(F'(x)\ne f(x)\), so \(F\) is not an antiderivative of \(f\).

No, \(F\) is not an antiderivative of \(f\).

- Rewrite the quotient using a negative exponent. - Apply the power rule for antiderivatives. - Different constants give different antiderivatives. - Substitute the given point to determine the constant.

1. Rewrite \(f_n(x)=3x^{-n}\). By the power rule, \(F_n(x)=\frac{3}{1-n}x^{1-n}+C=-\frac{3}{(n-1)x^{n-1}}+C\). 2. For each fixed \(n\), there are infinitely many antiderivatives because \(C\) can be any real number. 3. For \(n=4\), \(F_4(x)=-\frac1{x^3}+C\). Use \(F_4(1)=1\): \(-1+C=1\), so \(C=2\). Therefore, \(F_4(x)=-\frac1{x^3}+2\).

1. \(F_n(x)=-\frac{3}{(n-1)x^{n-1}}+C\), where \(C\in\mathbb R\) 2. Infinitely many 3. \(F_4(x)=-\frac1{x^3}+2\)

- The derivative of an antiderivative must equal the original function. - Use the quotient rule for rational functions with variable expressions in both numerator and denominator. - Rewrite reciprocal powers using negative exponents. - Simplify each derivative carefully before matching.

1. Differentiate \(F_1\) using the quotient rule: \(F_1'(x)=\frac{2(x+5)-2x}{(x+5)^2}=\frac{10}{(x+5)^2}=f_A(x)\). 2. Differentiate \(F_2\): \(F_2'(x)=\frac{2x(x-1)-x^2}{(x-1)^2}=\frac{x^2-2x}{(x-1)^2}=f_B(x)\). 3. Differentiate \(F_3(x)=4x^{-3}\): \(F_3'(x)=-12x^{-4}=f_C(x)\). 4. Differentiate \(F_4(x)=-x^{-4}\): \(F_4'(x)=4x^{-5}=\frac{4}{x^5}=f_D(x)\).

(A)–(1); (B)–(2); (C)–(3); (D)–(4)

- Recall the antiderivative of \(x^{-1}\). - Account for the sign of the logarithm's argument when \(x<0\). - Substitute each given function value into the general antiderivative to find \(C\). - Use basic logarithm values when simplifying.

1. On \(x<0\), the family of antiderivatives is \(F(x)=\frac{2}{3}\ln(-x)-4x+C\). 2. Use \(F_1(-1)=0\): \(\frac{2}{3}\ln(1)+4+C=0\). Since \(\ln(1)=0\), \(C=-4\). Thus, \(F_1(x)=\frac{2}{3}\ln(-x)-4x-4\). 3. Use \(F_2(-e)=1\): \(\frac{2}{3}\ln(e)+4e+C=1\). Therefore, \(\frac{2}{3}+4e+C=1\), so \(C=\frac{1}{3}-4e\). 4. Hence, \(F_2(x)=\frac{2}{3}\ln(-x)-4x+\frac{1}{3}-4e\).

1. \(F(x)=\frac{2}{3}\ln(-x)-4x+C\) 2. \(F_1(x)=\frac{2}{3}\ln(-x)-4x-4\) 3. \(F_2(x)=\frac{2}{3}\ln(-x)-4x+\frac{1}{3}-4e\)

- Differentiate \(G\) and compare it with \(g\). - Use the chain rule for \(\ln(-x)\). - For \(x>0\), integrate the reciprocal and polynomial terms. - Use the given function value to determine \(C\).

1. For \(x<0\), \(G'(x)=5\left(\frac{1}{-x}\right)(-1)+\frac{1}{2}x=\frac{5}{x}+\frac{1}{2}x=g(x)\). 2. For \(x>0\), the general antiderivative is \(H(x)=5\ln(x)+\frac{1}{4}x^2+C\). 3. Apply \(H(1)=0.25\): \(5\ln(1)+\frac{1}{4}+C=0.25\). 4. Since \(\ln(1)=0\) and \(\frac{1}{4}=0.25\), \(C=0\). Therefore, \(H(x)=5\ln(x)+\frac{1}{4}x^2\).

1. \(G'(x)=g(x)\) for \(x<0\). 2. \(H(x)=5\ln(x)+\frac{1}{4}x^2\)

- Check the derivative separately on each interval of the domain. - Use the chain rule for the logarithm in the region \(x<-4\). - Does a disconnected domain require the same additive constant on both components? - Compare the derivative with \(f\) on each interval.

1. For \(x>-4\), \(F'(x)=\frac{d}{dx}[\ln(x+4)-2]=\frac{1}{x+4}\). 2. For \(x<-4\), \(F'(x)=\frac{d}{dx}[\ln(-x-4)+5]=\frac{1}{-x-4}(-1)=\frac{1}{x+4}\). 3. Thus, \(F'(x)=f(x)\) on both components of the domain. 4. The different additive constants are allowed because the domain consists of two disconnected intervals. Therefore, \(F\) is an antiderivative of \(f\).

Yes. \(F^{\prime}(x)=f(x)\) on both \((-\infty, -4)\) and \((-4, \infty)\).

- Use the product rule. - Include the chain-rule factor when differentiating \(e^{2x}\). - Differentiate the candidates and compare each result with \(f\).

1. Differentiate \(F_1(x)=xe^{2x}+7\) using the product and chain rules. 2. \(F_1'(x)=e^{2x}+2xe^{2x}=(2x+1)e^{2x}=f(x)\). 3. The other derivatives are \(F_2'(x)=(2x+3)e^{2x}\), \(F_3'(x)=2e^{2x}\), and \(F_4'(x)=(2x+2)e^x\). 4. Therefore, only \(F_1\) is an antiderivative of \(f\).

\(F_1(x)=xe^{2x}+7\)

- Require the logarithm's input to be positive. - Use both the product rule and the chain rule when differentiating \((5-x)\ln(5-x)\). - Add a constant to the verified antiderivative and use the given point. - Recall the constant-difference property on a connected interval.

1. The logarithm requires \(5-x>0\), so the maximal domain is \((-\infty, 5)\). 2. Differentiate using the product and chain rules: \(G'(x)=3-\ln(5-x)-1=2-\ln(5-x)=g(x)\). Thus, \(G\) is an antiderivative of \(g\). 3. Write \(H(x)=G(x)+C\). Since \(H(4)=0\), \(0=3\cdot4+(5-4)\ln(1)+C=12+C\), so \(C=-12\). Therefore, \(H(x)=3x+(5-x)\ln(5-x)-12\). 4. The domain \((-\infty, 5)\) is one connected interval. On a connected interval, any two antiderivatives of the same function differ by a constant.

a) \((-\infty, 5)\) b) \(G'(x)=2-\ln(5-x)=g(x)\) c) \(H(x)=3x+(5-x)\ln(5-x)-12\) d) The domain is connected, so any two antiderivatives differ by one constant.

- Differentiate \(F\) to verify the antiderivative relationship. - Use the chain rule for a squared logarithm. - How do all antiderivatives of the same function differ? - Use the logarithm value at the given input when solving for \(C\).

1. Differentiate using the chain rule: \(F'(x)=\frac{1}{2}\cdot2\ln(x)\cdot\frac{1}{x}=\frac{\ln(x)}{x}=f(x)\). 2. Every antiderivative has the form \(G(x)=\frac{1}{2}[\ln(x)]^2+C\). 3. Use \(G(e)=0\): \(\frac{1}{2}[\ln(e)]^2+C=0\). 4. Since \(\ln(e)=1\), \(C=-\frac{1}{2}\). Thus, \(G(x)=\frac{1}{2}[\ln(x)]^2-\frac{1}{2}\).

1. \(F'(x)=\frac{\ln(x)}{x}=f(x)\). 2. \(G(x)=\frac{1}{2}[\ln(x)]^2-\frac{1}{2}\)

- Differentiate the proposed antiderivative. - Use the product rule on the term containing \(x^3\ln(x)\). - All antiderivatives differ by a constant. - Substitute the point to determine that constant.

1. Rewrite \(F(x)=\frac{1}{3}x^3\ln(x)-\frac{1}{9}x^3\). 2. Differentiate using the product rule: \(F'(x)=x^2\ln(x)+\frac{1}{3}x^2-\frac{1}{3}x^2=x^2\ln(x)=f(x)\). 3. Every antiderivative has the form \(H(x)=\frac{1}{9}x^3(3\ln(x)-1)+C\). 4. Use \(H(1)=1\): \(-\frac{1}{9}+C=1\), so \(C=\frac{10}{9}\). 5. Therefore, \(H(x)=\frac{1}{9}x^3(3\ln(x)-1)+\frac{10}{9}\).

1. \(F'(x)=x^2\ln(x)=f(x)\). 2. \(H(x)=\frac{1}{9}x^3(3\ln(x)-1)+\frac{10}{9}\)

- Pay attention to the logarithm's argument because \(x<0\). - Use the chain rule to check the derivative of \(\ln(-x)\). - Different constants give different antiderivatives.

1. Integrate term by term. An antiderivative of \(x\) is \(\frac12x^2\). 2. On \(x<0\), an antiderivative of \(\frac1x\) is \(\ln(-x)\). Therefore, a general antiderivative is \(F(x)=\frac12x^2-\frac14\ln(-x)+C\). 3. Choosing \(C=0\) and \(C=-1\) gives two different antiderivatives.

\(F_1(x)=\frac12x^2-\frac14\ln(-x)\) and \(F_2(x)=\frac12x^2-\frac14\ln(-x)-1\)

- Differentiate an antiderivative to recover the original function. - How do antiderivatives of the same function differ? - What happens when you differentiate their difference?

1. Since \(F\) is an antiderivative of \(f\), \(f(x)=F'(x)=2x^3-2x\). 2. Every antiderivative of \(f\) has the form \(G(x)=\frac{1}{2}x^4-x^2+C\). 3. Use \(G(2)=10\): \(8-4+C=10\), so \(C=6\). Thus, \(G(x)=\frac{1}{2}x^4-x^2+6\). 4. If \(F\) and \(G\) are both antiderivatives of \(f\), then \((G-F)'=G'-F'=f-f=0\). A function with derivative \(0\) on an interval is constant, so \(G-F=C\).

a) \(f(x)=2x^3-2x\) b) \(G(x)=\frac{1}{2}x^4-x^2+6\) c) The derivative of \(G-F\) is \(0\), so \(G-F\) is constant.

- First write the general antiderivative with a constant \(C\). - Substitute the given input and output values. - Solve each resulting equation for its constant.

1. The general antiderivative is \(F(x)=x^3+2x^2+C\). 2. From \(F_1(0)=0\), \(C_1=0\), so \(F_1(x)=x^3+2x^2\). 3. From \(F_2(0)=10\), \(C_2=10\), so \(F_2(x)=x^3+2x^2+10\). 4. From \(F_3(1)=5\), \(1+2+C_3=5\), so \(C_3=2\) and \(F_3(x)=x^3+2x^2+2\).

1. \(F_1(x)=x^3+2x^2\) 2. \(F_2(x)=x^3+2x^2+10\) 3. \(F_3(x)=x^3+2x^2+2\)

- Divide each term in the numerator by \(x^2\) before integrating. - Rewrite reciprocal powers using negative exponents. - Adding a constant to an antiderivative produces another antiderivative.

1. Simplify the function: \(f(x)=6x^2-3x^{-2}+5x^4\). 2. Integrating term by term gives an antiderivative \(F(x)=2x^3+3x^{-1}+x^5=x^5+2x^3+\frac3x\). 3. Adding different constants, such as \(0\) and \(5\), gives two different antiderivatives.

\(F_1(x)=x^5+2x^3+\frac3x\) \(F_2(x)=x^5+2x^3+\frac3x+5\)

- Expand the squared binomial first. - Rewrite the radical in the denominator using a negative exponent. - Integrate the resulting sum term by term. - Different constants produce different antiderivatives.

1. Expand and rewrite the function: \(g(x)=4x^4+4x^2+1-4x^{-1/2}\). 2. Integrating term by term gives \(G(x)=\frac45x^5+\frac43x^3+x-8\sqrt{x}+C\). 3. Choosing \(C=0\) and \(C=1\) gives two different antiderivatives.

\(G_1(x)=\frac45x^5+\frac43x^3+x-8\sqrt{x}\) \(G_2(x)=\frac45x^5+\frac43x^3+x-8\sqrt{x}+1\)

- Treat the parameter \(k\) as a constant. - Apply the reverse power rule to each term. - Differentiate your result to verify it.

1. Treat \(k\) as a constant and integrate each term with respect to \(x\). 2. The first term gives \(\frac{1}{k+1}\cdot\frac{x^{k+1}}{k+1}=\frac{x^{k+1}}{(k+1)^2}\). 3. The second term gives \(-k\cdot\frac{x^k}{k}=-x^k\). 4. Therefore, \(H(x)=\frac{x^{k+1}}{(k+1)^2}-x^k\).

\(H(x)=\frac{x^{k+1}}{(k+1)^2}-x^k\)

- Rewrite each function as a sum of powers before integrating. - Expand products and squared binomials when needed. - Rewrite terms with \(x\) in the denominator using negative exponents. - Treat \(p\) as a constant when integrating with respect to \(x\).

1. For a), expand to \(9x^2-6x+1\), then integrate term by term: \(F(x)=3x^3-3x^2+x\). 2. For b), expand to \(\frac12x^3+2x^2\), then integrate: \(F(x)=\frac18x^4+\frac23x^3\). 3. For c), simplify to \(x^2-2+6x^{-2}\), then integrate: \(F(x)=\frac13x^3-2x-\frac6x\). 4. For d), treat \(p\) as a constant. Expanding gives \(px^2-2p^2x+p^3\), so \(F(x)=\frac p3x^3-p^2x^2+p^3x\).

a) \(F(x)=3x^3-3x^2+x\) b) \(F(x)=\frac18x^4+\frac23x^3\) c) \(F(x)=\frac13x^3-2x-\frac6x\) d) \(F(x)=\frac p3x^3-p^2x^2+p^3x\)

- Expand powers of binomials before integrating. - Use exponent rules to simplify quotients. - Rewrite radicals and reciprocal powers using rational or negative exponents. - Differentiate each result to check it.

1. For a), expand to \(1-3x+3x^2-x^3\). Integrating term by term gives \(F(x)=x-\frac32x^2+x^3-\frac14x^4\). 2. For b), simplify to \(4x^3-2x^{-1/2}\). Integrating gives \(F(x)=x^4-4\sqrt{x}\). 3. For c), expand to \(x^4+4x^2+4\). Integrating gives \(F(x)=\frac15x^5+\frac43x^3+4x\). 4. For d), simplify to \(x-3x^{-2}+2x^{-4}\). Integrating gives \(F(x)=\frac12x^2+\frac3x-\frac{2}{3x^3}\).

a) \(F(x)=x-\frac32x^2+x^3-\frac14x^4\) b) \(F(x)=x^4-4\sqrt{x}\) c) \(F(x)=\frac15x^5+\frac43x^3+4x\) d) \(F(x)=\frac12x^2+\frac3x-\frac{2}{3x^3}\)

- First find the general antiderivative with a constant \(C\). - Substitute the coordinates of the given point. - Use \(\cos(0)=1\) to solve for \(C\).

1. The general antiderivative is \(F(x)=-\cos(x)-x^4+C\). 2. Use the condition \(F(0)=2\): \(-\cos(0)+C=2\). 3. Since \(\cos(0)=1\), \(-1+C=2\), so \(C=3\). 4. Therefore, \(F(x)=-\cos(x)-x^4+3\).

\(F(x)=-\cos(x)-x^4+3\)

- Account for the chain-rule factor when integrating \(e^{2x}\). - Evaluate the exponential term at the given input. - Use the given function value to find \(C\). - Differentiate your result to check it.

1. Integrate term by term: \(F(x)=2e^{2x}-2x^3+C\). 2. Apply \(F(0)=5\): \(2e^0+C=5\). 3. Thus, \(2+C=5\), so \(C=3\). 4. Therefore, \(F(x)=2e^{2x}-2x^3+3\).

\(F(x)=2e^{2x}-2x^3+3\)

- Integrate each polynomial term using the power rule. - What equation follows from a zero at \(x=2\)? - Include the integration constant in the general antiderivative. - Solve the resulting equation for \(C\).

1. The general antiderivative is \(F(x)=\frac{1}{8}x^4-x^3+5x+C\). 2. The zero at \(x=2\) means \(F(2)=0\): \(2-8+10+C=0\). 3. Therefore, \(4+C=0\), so \(C=-4\). 4. Thus, \(F(x)=\frac{1}{8}x^4-x^3+5x-4\).

\(F(x)=\frac{1}{8}x^4-x^3+5x-4\)

- Rewrite the reciprocal term using a negative exponent. - Apply the power rule for antiderivatives carefully. - Use the condition \(F(2)=0\). - Solve for the remaining constant.

1. Rewrite \(f(x)=4x^{-3}+\frac{1}{2}\). 2. Integrate: \(F(x)=-2x^{-2}+\frac{1}{2}x+C=-\frac{2}{x^2}+\frac{1}{2}x+C\). 3. Use \(F(2)=0\): \(-\frac{2}{4}+1+C=0\). 4. Thus, \(\frac{1}{2}+C=0\), so \(C=-\frac{1}{2}\). 5. Therefore, \(F(x)=-\frac{2}{x^2}+\frac{1}{2}x-\frac{1}{2}\).

\(F(x)=-\frac{2}{x^2}+\frac{1}{2}x-\frac{1}{2}\)

- How do antiderivatives of the same function differ? - Differentiate \(F\) to recover \(f\). - Use the point on \(G\) to determine the constant.

1. Recover \(f\) by differentiating \(F\): \(f(x)=F'(x)=-\frac{1}{x^2}+2x\). 2. Since \(F\) and \(G\) are antiderivatives of the same function, \(G(x)=F(x)+C=\frac{1}{x}+x^2+C\). 3. Use \(G(1)=5\): \(1+1+C=5\), so \(C=3\). 4. Therefore, \(G(x)=\frac{1}{x}+x^2+3\).

\(f(x)=-\frac{1}{x^2}+2x\) \(G(x)=\frac{1}{x}+x^2+3\)

- Recall the antiderivatives related to square-root and exponential functions. - Substitute the given input into the general antiderivative. - Solve the resulting equation for \(C\).

1. The general antiderivative is \(F(x)=\sqrt{x}+e^x+C\). 2. Apply \(F(1)=e+4\): \(1+e+C=e+4\). 3. Therefore, \(C=3\). 4. Thus, \(F(x)=\sqrt{x}+e^x+3\).

\(F(x)=\sqrt{x}+e^x+3\)

- First find the general antiderivative with a constant \(C\). - Account for the inner derivative of \(-x\). - Substitute the coordinates of the given point to determine \(C\).

1. Accounting for the inner derivative \(-1\), the general antiderivative is \(F(x)=-3e^{-x}+2x+C\). 2. Use \(F(0)=1\): \(-3e^0+C=1\). 3. Since \(e^0=1\), \(-3+C=1\), so \(C=4\). 4. Therefore, \(F(x)=-3e^{-x}+2x+4\).

\(F(x)=-3e^{-x}+2x+4\)

- Which logarithmic function differentiates to \(\frac{1}{x}\) when \(x<0\)? - Different antiderivatives differ only by constants. - How is the slope of an antiderivative related to \(f\)? - Evaluate \(f\) at the stated input.

1. On \(x<0\), the family of antiderivatives is \(F(x)=\ln(-x)+C\). 2. Three examples are \(F_1(x)=\ln(-x)\), \(F_2(x)=\ln(-x)+2\), and \(F_3(x)=\ln(-x)-1\). 3. Every antiderivative satisfies \(F'(x)=f(x)\), so the slope at \(x=-2\) is always \(f(-2)\). 4. Since \(f(-2)=-\frac{1}{2}\), all the antiderivatives have slope \(-\frac{1}{2}\) there.

a) \(F_1(x)=\ln(-x)\), \(F_2(x)=\ln(-x)+2\), and \(F_3(x)=\ln(-x)-1\) b) All have slope \(f(-2)=-\frac{1}{2}\).

- Read the slope and y-intercept of the line from the graph. - Integrate the derivative and include a constant \(C\). - Use the given point to determine \(C\).

1. The graph of \(f'\) is a line with y-intercept \(2\) and zero at \(x=4\). Its slope is \(\frac{0-2}{4-0}=-\frac{1}{2}\), so \(f'(x)=-\frac{1}{2}x+2\). 2. Integrate to obtain \(f(x)=-\frac{1}{4}x^2+2x+C\). 3. Use \(f(2)=1\): \(-1+4+C=1\), so \(C=-2\). 4. Therefore, \(f(x)=-\frac{1}{4}x^2+2x-2\).

\(f(x)=-\frac{1}{4}x^2+2x-2\)

- Match local extrema of a graph with zeros of its derivative. - Compare increasing and decreasing intervals with the sign of the derivative. - Use the polynomial degrees to check the order \(F\), \(f\), and \(f'\).

1. Graph B has a local maximum at \(x = 2\) and a local minimum at \(x = 6\). Graph C is zero at those values, is negative on \((2, 6)\), and is positive outside that interval. Therefore, graph C is the derivative of graph B. 2. Graph C has a local minimum at \(x = 4\). Graph A is zero at \(x = 4\), is negative for \(x < 4\), and is positive for \(x > 4\). Therefore, graph A is the derivative of graph C. 3. Since \(F' = f\) and \(f'\) is the derivative of \(f\), graph B represents \(F\), graph C represents \(f\), and graph A represents \(f'\).

A \(\rightarrow f'\); B \(\rightarrow F\); C \(\rightarrow f\)

- Differentiate the product \(x^2\ln(x)\) carefully and combine like terms. - Two antiderivatives of the same function differ by a constant; use \(H(1)=0\) to find it. - The sign of \(f=H'\) tells whether \(H\) is increasing or decreasing. - A change in \(H'\) from negative to positive identifies a minimum.

1. Differentiate using the product rule: \(F'(x)=x\ln(x)+\frac{1}{2}x-\frac{1}{2}x=x\ln(x)=f(x)\). Therefore, \(F\) is an antiderivative of \(f\). 2. Every antiderivative has the form \(H(x)=F(x)+C\). Since \(F(1)=-\frac{1}{4}\), the condition \(H(1)=0\) gives \(C=\frac{1}{4}\). Thus, \(H(x)=\frac{1}{2}x^2\ln(x)-\frac{1}{4}x^2+\frac{1}{4}\). 3. Because \(H'(x)=f(x)=x\ln(x)\), the sign of the displayed graph determines the monotonicity of \(H\). The graph is below the x-axis on \((0, 1)\) and above it on \((1, \infty)\). Therefore, \(H\) decreases on \((0, 1)\) and increases on \((1, \infty)\). 4. Since \(H'\) changes from negative to positive at \(x=1\), \(H\) has an absolute minimum at \(x=1\). The minimum point is \((1, 0)\).

a) \(F'(x)=x\ln(x)=f(x)\), so \(F\) is an antiderivative of \(f\). b) \(H(x)=\frac{1}{2}x^2\ln(x)-\frac{1}{4}x^2+\frac{1}{4}\) c) \(H\) decreases on \((0, 1)\), increases on \((1, \infty)\), and has an absolute minimum at \((1, 0)\).

- Use the intercepts and vertex of the parabola to determine \(f^{\prime}\). - Integrate \(f^{\prime}\) and include a constant of integration. - Use the point on \(f\) to determine the constant.

1. The derivative has zeros at \(x=-2\) and \(x=2\) and a y-intercept of \(4\). Therefore, \(f^{\prime}(x)=-(x-2)(x+2)=4-x^2\). 2. Integrating gives \(f(x)=4x-\frac13x^3+C\). 3. Use \(f(3)=2\): \(2=12-9+C\), so \(C=-1\). 4. Thus, \(f(x)=-\frac13x^3+4x-1\).

\(f(x)=-\frac13x^3+4x-1\)

- An antiderivative of a linear function is quadratic. - Use the zero of \(f\) to locate a horizontal tangent of \(F\). - Use \(F(2)=0\) to determine the vertical position of the graph.

1. The graph is the line \(f(x)=\frac12x-1\). 2. An antiderivative is \(F(x)=\frac14x^2-x+C\). 3. Use \(F(2)=0\): \(1-2+C=0\), so \(C=1\). 4. Therefore, \(F(x)=\frac14x^2-x+1=\frac14(x-2)^2\). Its graph is an upward-opening parabola with vertex \((2, 0)\).

\(F(x)=\frac14(x-2)^2\), an upward-opening parabola with vertex \((2, 0)\)

- The sign of \(f\) determines whether \(F\) increases or decreases. - The increasing or decreasing behavior of \(f\) determines the concavity of \(F\). - Use the initial condition to determine the constant of integration.

1. Since \(f(x)=\frac{2}{x+1}>0\) for \(x>-1\), every antiderivative is strictly increasing on that interval. 2. Integrating gives \(F(x)=2\ln(x+1)+C\). The condition \(F(0)=0\) gives \(C=0\), so \(F(x)=2\ln(x+1)\). 3. Because \(f\) is decreasing, the slope of \(F\) decreases, so \(F\) is concave down on \((-1, \infty)\). 4. As \(x\to-1^+\), \(F(x)\to-\infty\). As \(x\to\infty\), \(F\) continues to increase but becomes progressively flatter because \(f(x)\to0\).

\(F(x)=2\ln(x+1)\). Its graph passes through \((0, 0)\), is strictly increasing and concave down on \((-1, \infty)\), approaches \(-\infty\) as \(x\to-1^+\), and becomes flatter as \(x\to\infty\).

- Integrate \(x^{1/2}\) using the power rule. - Use \(F(0)=-1\) to determine the constant. - Use the sign and monotonicity of \(f\) to describe the shape of \(F\).

1. Integrating gives \(F(x)=\frac23x^{3/2}+C\). 2. The condition \(F(0)=-1\) gives \(C=-1\), so \(F(x)=\frac23x^{3/2}-1\). 3. Since \(f(x)>0\) for \(x>0\), \(F\) is strictly increasing there. Since \(f\) is increasing, the slope of \(F\) increases, so \(F\) is concave up. 4. The graph begins at \((0, -1)\) with a horizontal tangent. Useful points include \((1, -\frac13)\) and \((4, \frac{13}{3})\).

\(F(x)=\frac23x^{3/2}-1\). The graph starts at \((0, -1)\) with a horizontal tangent and is increasing and concave up for \(x>0\).

- Compare the y-intercept, x-intercepts, and horizontal end behavior of the four candidates. - Because \(F'=f\), decreasing and increasing parts of \(F\) correspond to negative and positive values of \(f\). - Use the quotient rule, then simplify the numerator before analyzing its sign.

1. The function satisfies \(F(0)=-2\), has zeros at \(x=-2\) and \(x=2\), and approaches \(1\) as \(x\to\pm\infty\). Only Graph 1 has all of these features. 2. Since \(F'=f\), the sign of \(f\) is determined by whether \(F\) is decreasing or increasing. Graph 1 is decreasing for \(x<0\), has a horizontal tangent at \(x=0\), and is increasing for \(x>0\). Therefore, \(f(x)<0\) for \(x<0\), \(f(0)=0\), and \(f(x)>0\) for \(x>0\). 3. Differentiate with the quotient rule: \(f(x)=F'(x)=\frac{2x(x^2+2)-2x(x^2-4)}{(x^2+2)^2}=\frac{12x}{(x^2+2)^2}\). 4. The denominator is positive for every real \(x\), so the sign of \(f\) is the sign of \(x\). This confirms the graph-based conclusions.

a) Graph 1 b) \(f(x)<0\) for \(x<0\), \(f(0)=0\), and \(f(x)>0\) for \(x>0\). c) \(f(x)=\frac{12x}{(x^2+2)^2}\), which has the stated signs because its denominator is always positive.

- Find the general antiderivative of \(\cos(kx)\). - Determine the range of \(\frac{1}{k}\sin(kx)\). - How does adding \(C\) shift the range vertically? - Set the minimum of the shifted range equal to \(0\).

1. The general antiderivative is \(F_k(x)=\frac{1}{k}\sin(kx)+C\). 2. Since \(\sin(kx)\) ranges from \(-1\) to \(1\), the term \(\frac{1}{k}\sin(kx)\) ranges from \(-\frac{1}{|k|}\) to \(\frac{1}{|k|}\). 3. Therefore, the minimum value of \(F_k\) is \(C-\frac{1}{|k|}\). 4. Set the minimum equal to \(0\): \(C-\frac{1}{|k|}=0\), so \(C=\frac{1}{|k|}\). 5. Thus \(F_k(x)=\frac{1}{k}\sin(kx)+\frac{1}{|k|}\).

\(F_k(x)=\frac{1}{k}\sin(kx)+\frac{1}{|k|}\)

- How are \(f\) and an antiderivative \(F\) related? - Where can \(F\) have local extrema? - How can the derivative of \(f\) distinguish minima from maxima of \(F\)? - How does the stated minimum value determine \(C\)?

1. The general antiderivative is \(F(x)=\frac{1}{4}x^4-2x^2+C\). 2. Critical points of \(F\) occur where \(F'(x)=f(x)=0\): \(x(x^2-4)=0\), so \(x=-2, 0, 2\). 3. Since \(F''(x)=f'(x)=3x^2-4\), the points at \(x=\pm2\) are local minima and the point at \(x=0\) is a local maximum. 4. Use the required minimum value: \(F(2)=5\). Then \(\frac{1}{4}(2)^4-2(2)^2+C=5\), so \(4-8+C=5\) and \(C=9\). 5. Therefore, \(F(x)=\frac{1}{4}x^4-2x^2+9\).

\(F(x)=\frac{1}{4}x^4-2x^2+9\)

- Which function has derivative \(\frac{1}{x}\)? - Set \(F'(x)=f(x)\) equal to zero to find critical points of \(F\). - Use logarithm properties when simplifying the condition. - The constant \(C\) shifts the graph vertically.

1. For \(x>0\), the general antiderivative is \(F(x)=\frac{1}{4}x^2-2\ln(x)+C\). 2. Critical points of \(F\) satisfy \(F'(x)=f(x)=0\): \(\frac{1}{2}x-\frac{2}{x}=0\), so \(x^2=4\). Because \(x>0\), \(x=2\). 3. Since \(F''(x)=f'(x)=\frac{1}{2}+\frac{2}{x^2}>0\), \(F\) has a local minimum at \(x=2\). 4. Apply the minimum value: \(F(2)=5-\ln(4)\). This gives \(1-2\ln(2)+C=5-\ln(4)\). 5. Because \(2\ln(2)=\ln(4)\), \(C=4\). Thus, \(F(x)=\frac{1}{4}x^2-2\ln(x)+4\).

\(F(x)=\frac{1}{4}x^2-2\ln(x)+4\)

- Which parts of the two terms are derivatives of each other? - Recall the relationship between \(\sqrt{x}\) and \(\frac{1}{2\sqrt{x}}\). - How does the negative sign relate to the derivative of cosine? - Try differentiating a product involving \(\sqrt{x}\) and \(\cos(x)\).

1. Look for a product whose derivative has the two given terms. 2. Let \(u(x)=\sqrt{x}\) and \(v(x)=\cos(x)\). Then \(u'(x)=\frac{1}{2\sqrt{x}}\) and \(v'(x)=-\sin(x)\). 3. By the product rule, \(\frac{d}{dx}[\sqrt{x}\cos(x)]=\frac{\cos(x)}{2\sqrt{x}}-\sqrt{x}\sin(x)=f(x)\). 4. Thus, one antiderivative is \(F(x)=\sqrt{x}\cos(x)\).

\(F(x)=\sqrt{x}\cos(x)\)

- Simplify \(f\) or differentiate the reciprocal term directly. - On each connected interval, antiderivatives differ by a constant. - Use the given point on the positive interval to determine one constant and the point on the negative interval to determine the other. - Consider the role of the excluded input \(x=0\).

1. Differentiate \(F\): \(F'(x)=\frac{1}{2}-\frac{2}{x^2}=\frac{x^2-4}{2x^2}=f(x)\). Thus, \(F\) is an antiderivative of \(f\) wherever \(x\ne0\). 2. On \((0, \infty)\), write \(G(x)=F(x)+C\). Since \(F(2)=1+4+1=6\), the condition \(G(2)=0\) gives \(C=-6\). Therefore, \(G(x)=\frac{1}{2}x-2+\frac{2}{x}\). 3. On \((-\infty, 0)\), write \(H(x)=F(x)+D\). Since \(F(-2)=-1+4-1=2\), the condition \(H(-2)=0\) gives \(D=-2\). Therefore, \(H(x)=\frac{1}{2}x+2+\frac{2}{x}\). 4. The domain is split into two disconnected intervals by \(x=0\). A value condition on one interval does not determine the vertical shift on the other interval, so the two constants may differ.

a) \(F'(x)=f(x)\) for \(x\ne0\). b) \(G(x)=\frac{1}{2}x-2+\frac{2}{x}\) for \(x>0\) c) \(H(x)=\frac{1}{2}x+2+\frac{2}{x}\) for \(x<0\) d) The two domain intervals are disconnected, so each interval can have its own constant of integration.

- Rewrite the reciprocal term as a negative power before differentiating. - Identify the connected intervals that make up the domain. - Determine one constant from \(A(1)=0\) and a separate constant from \(A(-1)=0\). - Verify the derivative on each branch of the piecewise function.

1. Differentiate \(g\): \(g'(x)=2+\frac{1}{x^2}=h(x)\), so \(g\) is an antiderivative on both intervals of the domain. 2. The domain \(x\ne0\) consists of the disconnected intervals \((-\infty, 0)\) and \((0, \infty)\). Antiderivatives may use one additive constant on the negative interval and a different additive constant on the positive interval. 3. Since \(g(1)=1\), the positive-domain formula needs constant \(-1\). Since \(g(-1)=-1\), the negative-domain formula needs constant \(1\). Thus, define \(A(x)=2x-\frac{1}{x}+1\) for \(x<0\), and \(A(x)=2x-\frac{1}{x}-1\) for \(x>0\). 4. Then \(A(-1)=0\) and \(A(1)=0\). Differentiating either branch gives \(2+\frac{1}{x^2}=h(x)\).

a) \(g'(x)=2+\frac{1}{x^2}=h(x)\). b) The domain has two disconnected intervals, so the additive constants may differ. c) \(A(x)=2x-\frac{1}{x}+1\) for \(x<0\), and \(A(x)=2x-\frac{1}{x}-1\) for \(x>0\).

- What differentiability condition must an antiderivative satisfy? - Recall the relationship between differentiability and continuity. - Compare the one-sided behavior and the defined value of \(G\) at \(x=0\). - Is satisfying the derivative equation away from one point enough on all of \(\mathbb{R}\)?

1. Check continuity of \(G\) at \(x=0\). The defined value is \(G(0)=2\), while \(\lim_{x\to0^-}G(x)=e^0=1\). 2. Since the left-hand limit does not equal the function value, \(G\) is discontinuous at \(x=0\). 3. A function must be differentiable at every point of an interval to be an antiderivative there, and differentiability implies continuity. 4. Although \(G'(x)=g(x)\) for \(x\ne0\), \(G\) is not differentiable at \(x=0\). Therefore, \(G\) is not an antiderivative of \(g\) on \(\mathbb{R}\).

No. \(G\) is discontinuous, and therefore not differentiable, at \(x=0\).

- Distinguish between a zero where the graph crosses the x-axis and a zero where it only touches. - Use the y-intercept to determine \(a\). - Integrate term by term and use the given point to find the constant. - A zero of \(f\) without a sign change gives \(F\) a horizontal tangent but not an extremum.

1. The graph touches the x-axis at \(x=2\), so \(x_1=2\) is a double zero. It crosses at \(x=-1\), so \(x_2=-1\). Since \(f(0)=-2\), \(-2=a(0-2)^2(0+1)=4a\), giving \(a=-\frac12\). Thus, \(f(x)=-\frac12(x-2)^2(x+1)=-\frac12x^3+\frac32x^2-2\). 2. Integrating gives \(F(x)=-\frac18x^4+\frac12x^3-2x+C\). The condition \(F(2)=0\) gives \(-2+4-4+C=0\), so \(C=2\). Therefore, \(F(x)=-\frac18x^4+\frac12x^3-2x+2\). 3. Since \(f(2)=F^{\prime}(2)=0\), the graph of \(F\) has a horizontal tangent at \(x=2\). The function \(f\) does not change sign there, so \(F\) does not have a local extremum. Because \(f\) has a local maximum at \(x=2\), \(F\) changes concavity there. Thus, \(F\) has a stationary inflection point at \(x=2\).

a) \(a=-\frac12\), \(x_1=2\), \(x_2=-1\); \(f(x)=-\frac12x^3+\frac32x^2-2\) b) \(F(x)=-\frac18x^4+\frac12x^3-2x+2\) c) A stationary inflection point at \(x=2\); \(f\) has a zero without a sign change and a local maximum there.