Free math worksheets

- Match the coefficient of \((x-1)^2\) with the Taylor coefficient formula. - Remember the factor of \(2!\).

1. In a Taylor polynomial centered at \(1\), the coefficient of \((x-1)^2\) is \(\frac{f''(1)}{2!}\). 2. Therefore, \(\frac{f''(1)}{2}=3\), so \(f''(1)=6\).

\(f''(1)=6\).

- First find the displacement from the center \(3\). - Substitute that displacement directly into the given polynomial.

1. Substitute \(x=3.1\), so \(x-3=0.1\). 2. Then \(P_2(3.1)=5+2(0.1)-(0.1)^2=5.19\).

\(f(3.1)\approx5.19\).

- Write the Taylor formula before substituting derivative values. - Divide the second-derivative value by \(2!\). - Evaluate using the displacement from the center, not the full input value.

1. The degree-\(2\) Taylor polynomial centered at \(2\) is \( P_2(x)=f(2)+f'(2)(x-2)+\frac{f''(2)}{2!}(x-2)^2\). 2. Substituting the given data gives \( P_2(x)=3-4(x-2)+3(x-2)^2\). 3. At \(x=2.05\), the change from the center is \(0.05\). Therefore, \( P_2(2.05)=3-4(0.05)+3(0.05)^2=2.8075\).

a) \( P_2(x)=3-4(x-2)+3(x-2)^2\). b) \(f(2.05)\approx2.8075\).

- Write the general Maclaurin formula through degree \(3\). - The first two coefficients look direct, but higher derivatives require factorial divisors. - Compare each term of the student's expression with the formula one degree at a time.

1. The degree-\(3\) Maclaurin formula is \( P_3(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3\). 2. Substitution gives \( P_3(x)=2+3x-2x^2+2x^3\). 3. The student used derivative values as coefficients without dividing the second- and third-derivative values by their factorials.

a) \( P_3(x)=2+3x-2x^2+2x^3\). b) The student omitted the factors \(2!\) and \(3!\) in the Taylor coefficients.

- Replace \(x\) with the center plus the displacement \(h\). - Expand before translating back to powers of \(x-1\). - Compare the degree of the function with the degree of the Taylor polynomial.

1. Since \(x=1+h\), substitute into the function: \( f(1+h)=(1+h)^3-2(1+h)\). 2. Expanding and combining terms gives \( f(1+h)=-1+h+3h^2+h^3\). 3. Replacing \(h\) by \(x-1\), the Taylor polynomial is \( P_3(x)=-1+(x-1)+3(x-1)^2+(x-1)^3\). 4. The original function is already a polynomial of degree \(3\), so its Taylor polynomial through degree \(3\) contains every term and is exact everywhere.

a) \( f(1+h)=-1+h+3h^2+h^3\). b) \( P_3(x)=-1+(x-1)+3(x-1)^2+(x-1)^3\). c) A degree-\(3\) polynomial is reproduced exactly by its third-degree Taylor polynomial.

- Multiply every term before combining like powers. - Cancellation occurs in the linear coefficient. - Truncation is based on total degree, not on which factor produced a term.

1. Multiplying gives \( P_2(x)Q_2(x)=1+\frac12x^2+\frac12x^3+\frac12x^4\). 2. Keeping terms through degree \(2\) gives the product approximation \( R_2(x)=1+\frac12x^2\). 3. At \(x=0.1\), \( R_2(0.1)=1+\frac12(0.1)^2=1.005\). 4. A degree-\(2\) Taylor model records only constant, linear, and quadratic behavior. Higher powers belong to the omitted remainder.

a) \( R_2(x)=1+\frac12x^2\). b) The product is approximately \(1.005\). c) Terms of degree \(3\) and higher lie beyond the requested Taylor degree.

- Match the value and first two derivatives of the exponential function at \(0\). - Read the sign of \(d\) by checking whether its graph lies above or below the x-axis. - Translate the sign of \(f-p\) into a comparison of \(f\) and \(p\).

1. For \(f(x)=e^x\), \(f(0)=1\), \(f'(0)=1\), and \(f''(0)=1\). Therefore, \(P_2(x)=1+x+\frac{x^2}{2}=p(x)\). 2. The graph places \(d(-0.5)\) below the x-axis and \(d(0.5)\) above the x-axis. Numerically, \(d(-0.5)=e^{-0.5}-0.625\approx-0.018469\) and \(d(0.5)=e^{0.5}-1.625\approx0.023721\). Thus \(p(-0.5)>f(-0.5)\), while \(f(0.5)>p(0.5)\). 3. The functions have the same value, first derivative, and second derivative at \(0\). Therefore, \(d(0)=d'(0)=d''(0)=0\), so the first possible nonzero local term has degree \(3\) or higher.

a) \(p(x)=1+x+\frac{x^2}{2}\) is the degree-\(2\) Maclaurin polynomial. b) \(d(-0.5)<0\), so \(p(-0.5)>f(-0.5)\). Also, \(d(0.5)>0\), so \(f(0.5)>p(0.5)\). c) The error function and its first two derivatives are \(0\) at the center.

- Integrate the polynomial term by term. - Increase each exponent by \(1\) and divide by the new exponent. - Apply the lower and upper limits only after finding the antiderivative.

1. Replace the integrand by its Taylor polynomial: \( \int_0^{0.2}\ln(1+x)\,dx\approx\int_0^{0.2}\left(x-\frac{x^2}{2}+\frac{x^3}{3}\right)dx\). 2. An antiderivative is \( \frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}\). 3. Evaluating at \(0.2\) gives \( \frac{(0.2)^2}{2}-\frac{(0.2)^3}{6}+\frac{(0.2)^4}{12}=0.0188\).

\( \int_0^{0.2}\ln(1+x)\,dx\approx0.0188\).

- A Taylor polynomial may be differentiated term by term. - Reduce each exponent by \(1\) and multiply by the original exponent. - Evaluate only after simplifying the derivative polynomial.

1. Differentiate the Taylor polynomial term by term: \( P_4'(x)=1-6x+\frac32x^2+4x^3\). 2. Substituting \(x=0.2\) gives \( P_4'(0.2)=1-6(0.2)+\frac32(0.2)^2+4(0.2)^3\). 3. The value is \(1-1.2+0.06+0.032=-0.108\).

a) \( P_4'(x)=1-6x+\frac32x^2+4x^3\). b) \(f'(0.2)\approx-0.108\).

- Substitute the displacement from the center before solving. - Keep the factor \(1/2\) attached to the unknown derivative. - Isolate the quadratic contribution after evaluating the constant and linear terms.

1. Substitute \(x=1.2\), so \(x-1=0.2\): \( 2.68=2+3(0.2)+\frac{f''(1)}{2}(0.2)^2\). 2. The known terms total \(2.6\), leaving \( 0.08=0.02f''(1)\). 3. Therefore, \(f''(1)=4\).

\(f''(1)=4\).

- Use the symmetry information before writing the polynomial. - Divide each even derivative by its factorial. - Evaluate the quadratic and quartic contributions separately.

1. Because \(f\) is even, every odd derivative at \(0\) is \(0\). 2. The polynomial is \( P_4(x)=2+\frac{-6}{2!}x^2+\frac{24}{4!}x^4=2-3x^2+x^4\). 3. The odd-power terms are absent because an even function has a Maclaurin series containing only even powers. 4. At \(x=0.3\), \( P_4(0.3)=2-3(0.3)^2+(0.3)^4=1.7381\).

a) \( P_4(x)=2-3x^2+x^4\). b) Even symmetry forces the odd derivatives at \(0\) to equal \(0\). c) \(f(0.3)\approx1.7381\).

- Substitute the entire inner expression into every occurrence of the polynomial variable. - Apply the power to the factor \(3\) as well as to \(x\). - Simplify the polynomial before evaluating it.

1. Replace the input \(u\) in the polynomial by \(3x\): \( Q_3(x)=1+2(3x)-(3x)^2+(3x)^3\). 2. Simplifying gives \( Q_3(x)=1+6x-9x^2+27x^3\). 3. At \(x=0.1\), \( Q_3(0.1)=1+0.6-0.09+0.027=1.537\).

a) \( Q_3(x)=1+6x-9x^2+27x^3\). b) \(g(0.1)\approx1.537\).

- Substitute the inner expression into the standard cosine series before truncating. - Simplify \((\sqrt{x})^{2n}\) carefully. - Match the desired cosine input to the square root of the polynomial input.

1. The cosine series is \( \cos u=1-\frac{u^2}{2!}+\frac{u^4}{4!}-\frac{u^6}{6!}+\cdots\). 2. Substitute \(u=\sqrt{x}\). Each even power of \(\sqrt{x}\) becomes an integer power of \(x\): \( \cos(\sqrt{x})=1-\frac{x}{2!}+\frac{x^2}{4!}-\frac{x^3}{6!}+\cdots\). 3. Therefore, \( P_3(x)=1-\frac{x}{2}+\frac{x^2}{24}-\frac{x^3}{720}\). 4. To approximate \(\cos(0.5)\), use \(x=(0.5)^2=0.25\): \( P_3(0.25)=1-\frac{0.25}{2}+\frac{0.25^2}{24}-\frac{0.25^3}{720}=\frac{40439}{46080}\approx0.877582465\).

a) \( P_3(x)=1-\frac{x}{2}+\frac{x^2}{24}-\frac{x^3}{720}\). b) \( \cos(0.5)\approx0.877582465\).

- Evaluate each function and its first derivative at the center. - Read the vertical separation at \(x=2\), then confirm it numerically. - Connect closeness near the center with the number of matching derivatives.

1. At \(x=0\), all four functions have value \(0\). Their derivatives at \(0\) are all \(1\), so they also share the same slope. 2. The graph shows that \(r\) is closest to \(f\) at \(x=2\). 3. Numerically, \(f(2)=\sin 2\approx0.909297427\), \(p(2)=2\), \(q(2)=\frac23\approx0.666666667\), and \(r(2)=\frac{14}{15}\approx0.933333333\). 4. The corresponding absolute errors are approximately \(1.090702573\), \(0.242630760\), and \(0.024035907\). Thus \(r\) is closest. 5. Each higher-degree Taylor polynomial matches more derivatives of \(\sin x\) at the center. Therefore, the difference begins with a higher power of \(x\), which is small near \(0\).

a) Shared value \(0\) and shared slope \(1\). b) \(r(x)\) is closest at \(x=2\). c) Higher-degree Taylor polynomials match more derivatives at the center, so their first possible error term has higher degree.

- Solve \(1+x=1.2\) before substituting. - Keep track of the alternating signs. - Compare the required value of \(x\) for \(\ln(4)\) with the given interval.

1. To approximate \(\ln(1.2)\), set \(1+x=1.2\), so \(x=0.2\). 2. The fourth-degree approximation is \(0.2-\frac{0.2^2}{2}+\frac{0.2^3}{3}-\frac{0.2^4}{4}=0.182266\ldots\). 3. Therefore, \(\ln(1.2)\approx0.1823\). 4. For \(\ln(4)\), direct substitution would require \(1+x=4\), or \(x=3\). Since \(3\) is outside the stated interval \((-1,1]\), the series representation does not apply there.

1. \(\ln(1.2)\approx0.1823\) 2. Direct substitution would require \(x=3\), which is outside \((-1,1]\).

- Compute the first three derivatives of the exponential function at \(0\). - Write the corresponding derivatives of a general cubic polynomial. - Evaluate the polynomial and its derivatives at \(0\). - Match corresponding values to determine the coefficients.

1. Compute the derivatives of \(f\): \(f'(x)=-2e^{-2x}\), \(f''(x)=4e^{-2x}\), and \(f'''(x)=-8e^{-2x}\). At \(x=0\), the values are \(f(0)=1\), \(f'(0)=-2\), \(f''(0)=4\), and \(f'''(0)=-8\). 2. For the cubic polynomial, \(p(0)=a_0\), \(p'(0)=a_1\), \(p''(0)=2a_2\), and \(p'''(0)=6a_3\). Matching the values gives \(a_0=1\), \(a_1=-2\), \(2a_2=4\), and \(6a_3=-8\). Therefore, \(a_2=2\) and \(a_3=-\frac{4}{3}\), so \(p(x)=-\frac{4}{3}x^3+2x^2-2x+1\).

\(p(x)=-\frac{4}{3}x^3+2x^2-2x+1\)

- Rewrite the input as the center plus a small displacement. - Use the angle-addition identity before substituting Maclaurin terms. - Keep only powers through degree \(3\) in the displacement.

1. Let \(h=x-\frac{\pi}{4}\). The angle-addition identity gives \( \sin\left(\frac{\pi}{4}+h\right)=\frac{\sqrt2}{2}(\cos h+\sin h)\). 2. Through degree \(3\), \( \cos h=1-\frac{h^2}{2}+\cdots\) and \( \sin h=h-\frac{h^3}{6}+\cdots\). 3. Therefore, \( P_3(x)=\frac{\sqrt2}{2}\left(1+h-\frac{h^2}{2}-\frac{h^3}{6}\right)\), where \(h=x-\frac{\pi}{4}\). 4. At \(h=0.1\), \( P_3\approx\frac{\sqrt2}{2}\left(1.1-0.005-0.000166667\right)\approx0.774164\).

\( P_3(x)=\frac{\sqrt2}{2}\left(1+h-\frac{h^2}{2}-\frac{h^3}{6}\right)\), where \(h=x-\frac{\pi}{4}\). \( \sin\left(\frac{\pi}{4}+0.1\right)\approx0.774164\).

- Replace the shifted expression with a single variable \(h\). - Solve the resulting quadratic equation. - Use the stated input interval to select the appropriate root.

1. Let \(h=x-3\). The model equation is \( 1+2h-h^2=1.75\). 2. Rearranging gives \(h^2-2h+0.75=0\). 3. Factoring gives \((h-0.5)(h-1.5)=0\), so \(h=0.5\) or \(h=1.5\). 4. The condition \(x\in[3, 4]\) means \(h\in[0, 1]\), so only \(h=0.5\) is allowed. 5. Therefore, the estimated input is \(x=3.5\).

\(x\approx3.5\).

- Rewrite the absolute-value function separately for positive and negative inputs. - Check derivatives at \(0\) using limits rather than only formulas valid away from \(0\). - A Taylor polynomial requires every derivative through its stated degree.

1. Clearly, \(f(0)=0\). 2. For \(x>0\), \(f(x)=x^3\); for \(x<0\), \(f(x)=-x^3\). Thus, for \(x\neq0\), \( f'(x)=3x|x|\). Also, \( f'(0)=\lim_{h\to0}\frac{|h|^3}{h}=0\). 3. Using the limit definition again, \( f''(0)=\lim_{h\to0}\frac{f'(h)-f'(0)}{h}=\lim_{h\to0}3|h|=0\). For \(x\neq0\), \(f''(x)=6|x|\). 4. The degree-\(2\) Maclaurin polynomial is \(P_2(x)=0\). 5. The one-sided derivatives of \(f''(x)=6|x|\) at \(0\) are \(-6\) and \(6\), so \(f'''(0)\) does not exist. Therefore, a degree-\(3\) Maclaurin polynomial is not defined.

a) \(f(0)=f'(0)=f''(0)=0\). b) \(P_2(x)=0\). c) \(f'''(0)\) does not exist because its one-sided values disagree.

- Work with the displacement \(h=x-2\) to simplify the quadratic. - Locate the vertex by setting the derivative of the model equal to \(0\). - Connect coefficient signs with slope and concavity at the center.

1. Let \(h=x-2\). Then \(P_2=5-2h+4h^2\). 2. The derivative with respect to \(h\) is \(-2+8h\). Setting it equal to \(0\) gives \(h=1/4\), so \(x=2.25\). 3. The minimum value is \( P_2(2.25)=5-2(0.25)+4(0.25)^2=4.75\). 4. The negative linear coefficient means the function initially decreases to the right of the center. The positive quadratic coefficient means the model curves upward, producing a local minimum.

a) \(x=2.25\). b) Minimum modeled output \(4.75\). c) The model initially decreases and is concave up near the center.

- Give the reciprocal polynomial one unknown coefficient for each power through \(x^2\). - In the product, the constant coefficient must be \(1\), while the coefficients of \(x\) and \(x^2\) must be \(0\). - Solve the coefficient equations in increasing order of degree.

1. Multiply \( (2+x-x^2)(a+bx+cx^2)\) and match coefficients through degree \(2\). 2. The constant equation is \(2a=1\), so \(a=1/2\). 3. The coefficient of \(x\) gives \(2b+a=0\), so \(b=-1/4\). 4. The coefficient of \(x^2\) gives \(2c+b-a=0\). Substituting \(a=1/2\) and \(b=-1/4\) gives \(c=3/8\). 5. Therefore, \( Q_2(x)=\frac12-\frac{x}{4}+\frac{3x^2}{8}\). 6. Multiplication gives \( P_2(x)Q_2(x)=1+\frac{5}{8}x^3-\frac{3}{8}x^4\), so the product equals \(1\) through degree \(2\), as required.

\( Q_2(x)=\frac12-\frac{x}{4}+\frac{3x^2}{8}\).

- Treat \(x+x^2\) as the input to the exponential polynomial. - When expanding powers of the input, discard terms above degree \(3\). - Combine like powers only after all three polynomial terms have been substituted.

1. Let \(u=x+x^2\). Through degree \(3\), \( e^u=1+u+\frac{u^2}{2}+\frac{u^3}{6}+\) terms of degree greater than \(3\). 2. Compute only the powers needed through degree \(3\): \( u^2=x^2+2x^3+\) higher-degree terms, and \( u^3=x^3+\) higher-degree terms. 3. Substitution gives \( P_3(x)=1+(x+x^2)+\frac{x^2+2x^3}{2}+\frac{x^3}{6}=1+x+\frac{3x^2}{2}+\frac{7x^3}{6}\). 4. Therefore, \( f(0.1)\approx1+0.1+\frac{3(0.1)^2}{2}+\frac{7(0.1)^3}{6}\approx1.116166667\).

\( P_3(x)=1+x+\frac{3x^2}{2}+\frac{7x^3}{6}\). \( f(0.1)\approx1.116166667\).

- Use displacements \(0.1\) and \(-0.1\) from the center. - Compare how even and odd powers behave when the sign of the displacement changes. - In the difference quotient, subtract before dividing by the full distance between the inputs.

1. At \(x=1.1\), the displacement is \(0.1\), so \( P_3(1.1)=5+0.2-0.03+0.004=5.174\). 2. At \(x=0.9\), the displacement is \(-0.1\), so \( P_3(0.9)=5-0.2-0.03-0.004=4.766\). 3. Their average is \( \frac{5.174+4.766}{2}=4.970\). The odd-power terms cancel in this average. 4. The symmetric difference quotient is \( \frac{5.174-4.766}{0.2}=2.04\). The constant and even-power terms cancel in the numerator of this quotient.

a) \(f(1.1)\approx5.174\) and \(f(0.9)\approx4.766\). b) Average \(4.970\); odd-power terms cancel. c) Symmetric difference quotient \(2.04\); constant and even-power terms cancel.

- Replace \(x\) with \(1+(x-1)\) before expanding. - Collect powers of the displacement only after expanding every term. - Distinguish evaluating a polynomial from obtaining a guaranteed approximation to the original function.

1. Let \(h=x-1\), so \(x=1+h\). Substitute into the polynomial: \( P_4(1+h)=2-(1+h)+\frac32(1+h)^2-\frac12(1+h)^3+\frac14(1+h)^4\). 2. Expanding and collecting powers of \(h\) gives \( P_4(1+h)=\frac94+\frac32h+\frac32h^2+\frac12h^3+\frac14h^4\). 3. Therefore, \( P_4(x)=\frac94+\frac32(x-1)+\frac32(x-1)^2+\frac12(x-1)^3+\frac14(x-1)^4\). 4. At \(x=1.08\), the displacement is \(h=0.08\), so \( P_4(1.08)=2.37986624\). 5. The original Maclaurin polynomial matches derivatives of \(f\) at \(0\). Rewriting the same polynomial around \(1\) neither proves derivative matching there nor supplies an error bound at \(1.08\). Thus the number in part b) is only a polynomial value, not a guaranteed approximation to \(f(1.08)\).

a) \( P_4(x)=\frac94+\frac32(x-1)+\frac32(x-1)^2+\frac12(x-1)^3+\frac14(x-1)^4\). b) \( P_4(1.08)=2.37986624\). c) Rewriting does not transfer Taylor matching to the new center and provides no remainder bound for \(f(1.08)\).

- Apply factorial scaling when converting coefficients to derivatives. - The lowest-degree nonzero term controls the local behavior near the center. - Comparing opposite inputs cancels every even-power contribution. - For part d), substitute \(x=0.1\) only after identifying which polynomial terms are nonzero.

1. Comparing with Taylor form gives \(f(0)=5\), \(f'(0)=f''(0)=f'''(0)=0\), \(f^{(4)}(0)=4!(-3)=-72\), and \(f^{(5)}(0)=5!(2)=240\). 2. Near \(0\), the first nonconstant term is \(-3x^4\). It is negative on both sides of \(0\) and has lower degree than the later term \(2x^5\). Therefore, \(f(x)<f(0)\) for all sufficiently small nonzero \(x\), so \(x=0\) is a strict local maximum. 3. The even fourth-degree terms cancel when opposite inputs are compared, while the fifth-degree terms give \( P_5(x)-P_5(-x)=4x^5\). Thus, for sufficiently small positive \(x\), \(f(x)>f(-x)\). 4. The approximation is \( f(0.1)\approx5-3(0.1)^4+2(0.1)^5=4.99972\).

a) \(f(0)=5\), \(f'(0)=f''(0)=f'''(0)=0\), \(f^{(4)}(0)=-72\), and \(f^{(5)}(0)=240\). b) \(x=0\) is a strict local maximum. c) For small \(x>0\), \(f(x)>f(-x)\). d) \(f(0.1)\approx4.99972\).

- Connect the missing quadratic coefficient to the second derivative at the center. - Determine the center before substituting a shifted variable. - Remember that the coefficient of \((x-a)^n\) equals \(f^{(n)}(a)/n!\). - For part d), use the centered form and the displacement from the recovered center.

1. A missing quadratic term means the second derivative of the polynomial is \(0\) at the center. 2. Since \( P''(x)=12x-24\), solving \(P''(a)=0\) gives \(a=2\). 3. Let \(h=x-2\). Substituting \(x=2+h\) and collecting powers gives \( P(x)=3+(x-2)+2(x-2)^3\). 4. Comparing with Taylor form gives \(f(2)=3\), \(f'(2)=1\), \(f''(2)=0\), and \( \frac{f'''(2)}{3!}=2\), so \(f'''(2)=12\). 5. At \(x=2.1\), the displacement is \(0.1\). Therefore, \( f(2.1)\approx3+0.1+2(0.1)^3=3.102\).

a) \(a=2\). b) \( P(x)=3+(x-2)+2(x-2)^3\). c) \(f(2)=3\), \(f'(2)=1\), \(f''(2)=0\), and \(f'''(2)=12\). d) \(f(2.1)\approx3.102\).

- Differences of values at \(x\) and \(-x\) isolate odd powers. - Averages of values at opposite inputs isolate even powers. - Apply factorial scaling only after all polynomial coefficients are known.

1. Write \( P_4(x)=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4\). Since \(P_4(0)=3\), \(c_0=3\). 2. Opposite-input differences isolate the odd coefficients: \( \frac{P_4(1)-P_4(-1)}{2}=c_1+c_3=0\), and \( \frac{P_4(2)-P_4(-2)}{4}=c_1+4c_3=-3\). Thus \(c_3=-1\) and \(c_1=1\). 3. Opposite-input averages isolate the even coefficients: \(c_2+c_4=3\) and \(c_2+4c_4=6\). Hence \(c_4=1\) and \(c_2=2\). 4. Therefore, \( P_4(x)=3+x+2x^2-x^3+x^4\). 5. Comparing with Taylor form gives \(f(0)=3\), \(f'(0)=1\), \(f''(0)=4\), \(f'''(0)=-6\), and \(f^{(4)}(0)=24\). 6. Finally, \( f(0.3)\approx P_4(0.3)=3.4611\).

a) \( P_4(x)=3+x+2x^2-x^3+x^4\). b) \(f(0)=3\), \(f'(0)=1\), \(f''(0)=4\), \(f'''(0)=-6\), and \(f^{(4)}(0)=24\). c) \(f(0.3)\approx3.4611\).

- List exactly which derivatives of \(f\) are determined by a degree-\(3\) Maclaurin polynomial. - Multiply the known polynomial and keep only coefficients that cannot involve an unknown fourth derivative. - Use the family \(P_3(x)+cx^4\) to show that the next coefficient can vary. - In part d), identify the first degree produced by multiplying an unknown fourth-degree term by the nonzero linear term.

1. A degree-\(3\) Maclaurin polynomial determines \(f(0)\), \(f'(0)\), \(f''(0)\), and \(f'''(0)\), but it gives no information about \(f^{(4)}(0)\). 2. The derivatives of \(f^2\) through order \(3\) depend only on those known derivatives of \(f\). Equivalently, multiplying the known coefficients and retaining powers through degree \(3\) gives \( [P_3(x)]^2=1+2x+5x^2+2x^3+\text{higher-degree terms}\). Therefore, the guaranteed Maclaurin polynomial is \(1+2x+5x^2+2x^3\). 3. Every function \(f_c(x)=P_3(x)+cx^4\) has the same derivatives through order \(3\) at \(0\). The coefficient of \(x^4\) in \([P_3(x)]^2\) is \(2\), and the cross term \(2P_3(x)(cx^4)\) contributes \(2c\) because \(P_3(0)=1\). Thus the possible \(x^4\) coefficient is \(2+2c\), so it is not determined. 4. In the alternate case, the known polynomial begins with a nonzero linear term. The unknown fourth-degree behavior first combines with that linear term to affect degree \(5\). Therefore, the coefficients of \(f^2\) through degree \(4\) are guaranteed.

a) The degree-\(3\) Maclaurin polynomial is the highest one guaranteed. b) \(1+2x+5x^2+2x^3\). c) The coefficient of \(x^4\) in \(f_c(x)^2\) is \(2+2c\), so it is not fixed by the given polynomial. d) The Maclaurin polynomial for \(f^2\) is guaranteed through degree \(4\).

- Expand only the powers of \(x+x^2\) that can contribute through degree \(4\). - Match coefficients from lowest degree to highest; each equation introduces one new unknown. - For the last part, compare the lowest power of \(x\) in \((x+x^2)^k\) with the value of \(k\). - In part d), note that \((x+x^2)^k\) has lowest degree \(k\).

1. The powers needed through degree \(4\) are \(u=x+x^2\), \(u^2=x^2+2x^3+x^4\), \(u^3=x^3+3x^4+O(x^5)\), and \(u^4=x^4+O(x^5)\). 2. Substitution gives \(P_4(x+x^2)=a_0+a_1x+(a_1+a_2)x^2+(2a_2+a_3)x^3+(a_2+3a_3+a_4)x^4+O(x^5)\). 3. Matching coefficients with \(Q_4\) produces \(a_0=1\), \(a_1=2\), \(a_1+a_2=5\), \(2a_2+a_3=10\), and \(a_2+3a_3+a_4=20\). Solving in order gives \(a_2=3\), \(a_3=4\), and \(a_4=5\). 4. Therefore, \(P_4(u)=1+2u+3u^2+4u^3+5u^4\). At \(u=0.1\), \(P_4(0.1)=1+0.2+0.03+0.004+0.0005=1.2345\). 5. Every omitted term of the series for \(f\) contains \(u^k\) with \(k\geq5\). Because \(u=x+x^2\) has lowest degree \(1\), the lowest degree in \(u^k\) is \(x^k\). Such terms begin at degree \(5\) and cannot affect coefficients through degree \(4\).

a) \(a_0=1\), \(a_1=2\), \(a_1+a_2=5\), \(2a_2+a_3=10\), and \(a_2+3a_3+a_4=20\). b) \(P_4(u)=1+2u+3u^2+4u^3+5u^4\). c) \(f(0.1)\approx1.2345\) using the degree-\(4\) Maclaurin polynomial. d) A term \(u^k\) with \(k\geq5\) begins with \(x^k\) after substituting \(u=x+x^2\), so it contributes no term below degree \(5\).