Free math worksheets

- Match the inverse input with the original input-output pair. - At corresponding points, inverse derivative values are reciprocals when the original derivative is nonzero. - Use the derivative value attached to the original input.

1. Since \(f(2)=7\), the corresponding inverse input is \(7\). 2. Use the inverse-derivative relationship: \(\left(f^{-1}\right)^{\prime}(7)=\frac{1}{f^{\prime}(2)}\). 3. Therefore, \(\left(f^{-1}\right)^{\prime}(7)=\frac{1}{4}\).

\(\frac{1}{4}\)

- Choose two clear grid-intersection points on each line. - Compute each line's slope as rise divided by run. - Compare the two slopes after remembering that inverse graphs reflect across \(y=x\).

1. From the grid, the line \(f\) rises \(2\) units for each \(1\)-unit run, so \(f^{\prime}(x)=2\). 2. The inverse line \(g\) rises \(1\) unit for each \(2\)-unit run, so \(g^{\prime}(x)=\frac{1}{2}\). 3. The slopes are reciprocals, as expected for inverse functions at corresponding points.

\(f^{\prime}(x)=2\) and \(g^{\prime}(x)=\frac{1}{2}\)

- Read an inverse value by reversing the input-output relationship in the table. - For an inverse derivative, first locate the original input that maps to the requested inverse input. - Use the derivative value from the same row as that original input.

1. a) The table shows \(f(2)=7\), so \(f^{-1}(7)=2\). 2. b) The input that produces \(3\) is \(0\), so use the inverse-derivative formula at the corresponding point: \(\left(f^{-1}\right)^{\prime}(3)=\frac{1}{f^{\prime}(0)}=\frac{1}{2}\).

a) \(2\) b) \(\frac{1}{2}\)

- A function that is strictly monotonic is one-to-one. - The natural logarithm reverses an exponential function. - The domain of the inverse is the range of the original function. - At corresponding points, the derivative values of a function and its inverse are reciprocals.

1. Since \(g'(x)=\frac{1}{2}e^{x-3}>0\) for every real \(x\), the function is strictly increasing and therefore one-to-one. Thus, it has an inverse on its entire domain. 2. From \(y=\frac{1}{2}e^{x-3}+1\), obtain \(2(y-1)=e^{x-3}\). Taking natural logarithms gives \(x=\ln(2y-2)+3\). 3. Therefore, \(g^{-1}(x)=\ln(2x-2)+3\). Its domain is \((1,\infty)\), because \(2x-2>0\). 4. Since \(g'(3)=\frac{1}{2}\), the inverse-function derivative rule gives \(\left(g^{-1}\right)'(1.5)=\frac{1}{g'(3)}=2\).

1) \(g'(x)>0\) for all real \(x\), so \(g\) is strictly increasing and invertible. 2) \(g^{-1}(x)=\ln(2x-2)+3\), with domain \((1,\infty)\) 3) \(2\)

- Interchange the input and output, then solve while respecting the restricted domain. - Examine the sign of \(f(x)-x\) only where \(f\) is defined. - Inverse graphs are reflections across \(y=x\). - At corresponding points, inverse-function derivative values are reciprocals.

1. From \(y=\frac{1}{2}x^2+1\), obtain \(x^2=2y-2\). Because the original domain requires \(x\ge0\), \(x=\sqrt{2y-2}\). 2. Thus, \(f^{-1}(x)=\sqrt{2x-2}\). The range of \(f\) is \([1,\infty)\), so the inverse domain is \([1,\infty)\). 3. For every \(x\) in the domain \([0,\infty)\), \(f(x)-x=\frac{1}{2}x^2-x+1=\frac{1}{2}(x-1)^2+\frac{1}{2}>0\). Therefore, the graph of \(f\) lies above \(y=x\) and never intersects it. 4. Since inverse graphs are reflections across \(y=x\), the graph of \(f^{-1}\) lies below \(y=x\). The two graphs lie on opposite sides of the line and do not intersect. 5. Since \(f^{\prime}(x)=x\), \(f^{\prime}(2)=2\), and \(f(2)=3\). Therefore, \(\left(f^{-1}\right)^{\prime}(3)=\frac{1}{f^{\prime}(2)}=\frac{1}{2}\).

1) \(f^{-1}(x)=\sqrt{2x-2}\), with domain \([1,\infty)\) 2) The graph of \(f\) lies above \(y=x\), and the graph of \(f^{-1}\) lies below \(y=x\); they do not intersect. 3) The slopes are \(2\) for \(f\) at \(x=2\) and \(\frac{1}{2}\) for \(f^{-1}\) at \(x=3\).

- Apply the chain rule to the composition \(f(g(x))\). - The derivative of the identity function is \(1\). - The exponential function is its own derivative. - Use \(e^{\ln(x)}=x\).

1. a) Differentiate \(f(g(x))=x\). By the chain rule, \(f'(g(x))g'(x)=1\). 2. Provided \(f'(g(x))\ne0\), solve for the inverse derivative: \(g'(x)=\frac{1}{f'(g(x))}\). 3. b) For \(f(x)=e^x\), \(f'(x)=e^x\). Since \(g(x)=\ln(x)\), the formula gives \(g'(x)=\frac{1}{e^{g(x)}}=\frac{1}{e^{\ln(x)}}\). 4. Therefore, \(g'(x)=\frac{1}{x}\) for \(x>0\).

a) \(g'(x)=\frac{1}{f'(g(x))}\), where \(f'(g(x))\ne0\) b) \(\frac{d}{dx}\ln(x)=\frac{1}{x}\) for \(x>0\)

- A positive first derivative implies strict increase. - Try a small integer input that makes \(f(x)=2\). - At corresponding points, derivative values of a function and its inverse are reciprocals. - Evaluate the original derivative at the preimage of \(2\).

1. a) \(f'(x)=3x^2+2\ge2>0\) for every real \(x\). Therefore, \(f\) is strictly increasing and has an inverse function on \(\mathbb{R}\). 2. b) Testing \(a=1\) gives \(f(1)=1+2-1=2\), so \(f^{-1}(2)=1\). 3. By the inverse-function derivative rule, \(\left(f^{-1}\right)'(2)=\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(1)}\). 4. Since \(f'(1)=3(1)^2+2=5\), \(\left(f^{-1}\right)'(2)=\frac{1}{5}\).

a) \(f\) is invertible because \(f'(x)>0\) for all real \(x\). b) \(\left(f^{-1}\right)'(2)=\frac{1}{5}\)

- The restricted domain determines which square-root branch to use. - Evaluate the inverse first to find the corresponding original input. - At corresponding points, nonzero derivative values are reciprocals.

1. a) Start with \(y=2x^2+4\). Then \(x^2=\frac{y-4}{2}\). 2. Because the original domain requires \(x\le0\), take the negative square root: \(x=-\sqrt{\frac{y-4}{2}}\). 3. Therefore, \(g^{-1}(x)=-\sqrt{\frac{x-4}{2}}\). 4. b) First, \(g^{-1}(12)=-\sqrt{\frac{8}{2}}=-2\). 5. Since \(g'(x)=4x\), the inverse derivative is \(\left(g^{-1}\right)'(12)=\frac{1}{g'(-2)}=\frac{1}{-8}=-\frac{1}{8}\).

a) \(g^{-1}(x)=-\sqrt{\frac{x-4}{2}}\) b) \(-\frac{1}{8}\)

- Inverse functions interchange point coordinates. - Use \(f(-x)=-f(x)\) and the defining relationship between a function and its inverse. - At corresponding points, nonzero derivative values are reciprocals.

1. a) Inverse functions interchange input and output coordinates, so the corresponding point is \((8,3)\). 2. b) Origin symmetry means \(f(-x)=-f(x)\). Let \(y=f^{-1}(x)\), so \(f(y)=x\). 3. Then \(f(-y)=-f(y)=-x\). Applying \(f^{-1}\) gives \(-y=f^{-1}(-x)\). 4. Since \(y=f^{-1}(x)\), this becomes \(f^{-1}(-x)=-f^{-1}(x)\). Therefore, \(f^{-1}\) is odd and its graph has origin symmetry. 5. c) Since \(f(1)=4\) and \(f'(1)=5\), the inverse-function derivative rule gives \(\left(f^{-1}\right)'(4)=\frac{1}{f'(1)}=\frac{1}{5}\).

a) \((8,3)\) b) \(f^{-1}(-x)=-f^{-1}(x)\), so the inverse graph has origin symmetry. c) \(\left(f^{-1}\right)'(4)=\frac{1}{5}\)

- Use the derivative to show strict increase. - The domain restriction determines the square-root sign. - Evaluate the inverse to find the corresponding original input. - At corresponding points, nonzero derivative values are reciprocals.

1. a) \(g'(x)=x-3\). For \(x>3\), the derivative is positive, so \(g\) is strictly increasing on \([3,\infty)\) and is invertible. 2. Since \(g(3)=0\) and \(g\) increases without bound, its range is \([0,\infty)\). 3. b) Start with \(y=\frac{1}{2}(x-3)^2\). Since \(x\ge3\), \(x-3=\sqrt{2y}\), so \(x=\sqrt{2y}+3\). 4. Therefore, \(g^{-1}(x)=\sqrt{2x}+3\). 5. c) \(g^{-1}(2)=5\), and \(g'(5)=2\). By the inverse-function derivative rule, \(\left(g^{-1}\right)'(2)=\frac{1}{g'(5)}=\frac{1}{2}\).

a) \(g\) is strictly increasing; range: \([0,\infty)\) b) \(g^{-1}(x)=\sqrt{2x}+3\) c) \(\frac{1}{2}\)

- Solve the original equation for the input, then interchange the variables. - Because the function is strictly increasing, a shared point with its inverse lies on \(y=x\). - Differentiate the radical and quadratic functions. - Multiply the two tangent slopes to test the reciprocal relationship.

1. Write \(y=2\sqrt{x}\). Then \(\frac{y}{2}=\sqrt{x}\), so \(x=\frac{y^2}{4}\). Interchanging the variables gives \(f^{-1}(x)=\frac{x^2}{4}\). 2. The range of \(f\) is \([0,\infty)\), so the domain of \(f^{-1}\) is \([0,\infty)\). 3. At an intersection, \(2\sqrt{x}=\frac{x^2}{4}\). For \(x\ge0\), squaring gives \(4x=\frac{x^4}{16}\), so \(x(x^3-64)=0\). Besides \(x=0\), \(x=4\), giving the point \((4,4)\). 4. The derivatives are \(f^{\prime}(x)=\frac{1}{\sqrt{x}}\) and \((f^{-1})^{\prime}(x)=\frac{x}{2}\). 5. At \(x=4\), \(f^{\prime}(4)=\frac{1}{2}\) and \((f^{-1})^{\prime}(4)=2\). Their product is \(1\), so the slopes are reciprocals.

a) \(f^{-1}(x)=\frac{x^2}{4}\), with domain \([0,\infty)\) b) \((4,4)\) c) \(f^{\prime}(4)=\frac{1}{2}\) and \((f^{-1})^{\prime}(4)=2\), so the slopes are reciprocals.

- Differentiate the original function before using any inverse-function formula. - Check the denominator condition in the reciprocal derivative relationship. - Reflect the tangent line to \(f\) across \(y=x\) to infer the tangent to its inverse.

1. a) Since \(f^{\prime}(x)=3x^2\), \(f^{\prime}(0)=0\). 2. b) The inverse-derivative formula would require \(g^{\prime}(0)=\frac{1}{f^{\prime}(0)}\), but the denominator is \(0\). Therefore, the formula does not give a finite derivative. 3. c) The tangent to \(f\) at \(P=(0,0)\) is horizontal, with equation \(y=0\). Reflecting this tangent across \(y=x\) gives the vertical line \(x=0\). Thus, the inverse curve has a vertical tangent at \(P\).

a) \(0\) b) The reciprocal formula would divide by \(0\), so \(g^{\prime}(0)\) is not a finite derivative value. c) \(x=0\)

- The derivative gives the tangent slope. - At corresponding inverse points, nonzero tangent slopes are reciprocals. - Use point-slope form for each tangent line. - Apply the formula for the acute angle between two lines.

1. The derivative is \(f'(x)=\frac{2}{3}x\), so the slope of the tangent to \(f\) at \(x=1\) is \(m_1=\frac{2}{3}\). 2. Using point-slope form, \(y-1=\frac{2}{3}(x-1)\), so the tangent to \(f\) is \(y=\frac{2}{3}x+\frac{1}{3}\). 3. At the corresponding point of the inverse, the slope is reciprocal: \(m_2=\frac{1}{f'(1)}=\frac{3}{2}\). 4. Thus, \(y-1=\frac{3}{2}(x-1)\), so the tangent to \(f^{-1}\) is \(y=\frac{3}{2}x-\frac{1}{2}\). 5. The acute angle between lines with slopes \(m_1\) and \(m_2\) satisfies \(\tan(\alpha)=\left|\frac{m_2-m_1}{1+m_1m_2}\right|\). 6. Therefore, \(\tan(\alpha)=\frac{\frac{3}{2}-\frac{2}{3}}{1+\frac{2}{3}\cdot\frac{3}{2}}=\frac{5}{12}\), so \(\alpha=\arctan\!\left(\frac{5}{12}\right)\approx22.62^\circ\).

Tangent to \(f\): \(y=\frac{2}{3}x+\frac{1}{3}\) Tangent to \(f^{-1}\): \(y=\frac{3}{2}x-\frac{1}{2}\) \(\alpha\approx22.62^\circ\)

- For a strictly increasing function, shared points with the inverse lie on \(y=x\). - Use reciprocal slopes at corresponding inverse points. - Apply the formula for the acute angle between two lines.

1. a) Because \(f\) is strictly increasing, its intersections with its inverse lie on \(y=x\). Solve \(x^3=x\). On \([0,\infty)\), the nonzero solution is \(x=1\), so \(S=(1,1)\). 2. b) Since \(f'(x)=3x^2\), the tangent slope to \(f\) at \(S\) is \(3\). Thus, \(y-1=3(x-1)\), or \(y=3x-2\). 3. The inverse tangent slope is \(\frac{1}{3}\). Thus, \(y-1=\frac{1}{3}(x-1)\), or \(y=\frac{1}{3}x+\frac{2}{3}\). 4. c) The acute angle satisfies \(\tan(\alpha)=\left|\frac{3-\frac{1}{3}}{1+3\cdot\frac{1}{3}}\right|=\frac{4}{3}\). 5. Therefore, \(\alpha=\arctan\!\left(\frac{4}{3}\right)\approx53.13^\circ\).

a) \(S=(1,1)\) b) Tangent to \(f\): \(y=3x-2\); tangent to \(f^{-1}\): \(y=\frac{1}{3}x+\frac{2}{3}\) c) \(\alpha\approx53.13^\circ\)

- Study the sign of the first derivative for different values of \(a\). - Combine the logarithms when subtracting the two x-coordinates. - Substitute \(u=e^x\) when solving for the inverse. - Differentiate the inverse function with the chain rule.

1. The first derivative is \(f_a'(x)=e^x(2e^x+a)\). For \(a\ge0\), it is positive for every \(x\), so the function is strictly increasing, has no extrema, and is one-to-one on \(\mathbb{R}\). For \(a<0\), there is a local minimum at \(x=\ln\left(-\frac{a}{2}\right)\), so the function is not one-to-one on all of \(\mathbb{R}\). 2. The second derivative is \(f_a''(x)=e^x(4e^x+a)\). For \(a<0\), it is zero at \(x=\ln\left(-\frac{a}{4}\right)\) and changes from negative to positive there, so this is the inflection-point x-coordinate. Therefore, the horizontal distance is \(\ln\left(-\frac{a}{2}\right)-\ln\left(-\frac{a}{4}\right)=\ln(2)\). 3. For \(a=2\), write \(y=e^{2x}+2e^x=(e^x+1)^2-1\). Since \(e^x>0\), \(e^x=\sqrt{y+1}-1\). Thus, \(f_2^{-1}(x)=\ln(\sqrt{x+1}-1)\), with domain \((0,\infty)\). 4. Differentiating gives \((f_2^{-1})'(x)=\frac{1}{2(x+1-\sqrt{x+1})}\).

a) For \(a\ge0\), \(f_a\) is strictly increasing and one-to-one on \(\mathbb{R}\). For \(a<0\), it has a local minimum and is not one-to-one on \(\mathbb{R}\). b) The horizontal distance is \(\ln(2)\). c) \(f_2^{-1}(x)=\ln(\sqrt{x+1}-1)\), domain \((0,\infty)\) d) \((f_2^{-1})'(x)=\frac{1}{2(x+1-\sqrt{x+1})}\)

- Use the behavior of \(e^x\) for the limits. - Factor the second derivative completely. - A derivative that is nonnegative and zero only at an isolated point can still correspond to a strictly increasing function. - Use the inverse-derivative formula after finding the corresponding original input.

1. The zero satisfies \(e^x=k\), so \(x=\ln(k)\). As \(x\to\infty\), \(g_k(x)\to\infty\). As \(x\to-\infty\), \(e^x\to0\), so \(g_k(x)\to-k^3\). 2. The second derivative is \(g_k''(x)=3e^x(e^x-k)(3e^x-k)\). Its zeros are \(x=\ln(k)\) and \(x=\ln\left(\frac{k}{3}\right)\), and the second derivative changes sign at both. Their horizontal distance is \(\ln(k)-\ln\left(\frac{k}{3}\right)=\ln(3)\). 3. The first derivative is \(g_k'(x)=3(e^x-k)^2e^x\ge0\), with equality only at one isolated point. Therefore, \(g_k\) is strictly increasing and one-to-one. 4. Solving \(y=(e^x-k)^3\) gives \(g_k^{-1}(x)=\ln(\sqrt[3]{x}+k)\). Its domain is \((-k^3,\infty)\). 5. The input \(k^3\) corresponds to \(e^x=2k\), or \(x=\ln(2k)\). Since \(g_k'(\ln(2k))=6k^3\), the inverse-derivative formula gives \((g_k^{-1})'(k^3)=\frac{1}{6k^3}\).

a) Zero: \(x=\ln(k)\); \(g_k(x)\to\infty\) as \(x\to\infty\); \(g_k(x)\to-k^3\) as \(x\to-\infty\) b) Inflection x-coordinates: \(\ln(k)\) and \(\ln\left(\frac{k}{3}\right)\); distance \(\ln(3)\) c) \(g_k^{-1}(x)=\ln(\sqrt[3]{x}+k)\), domain \((-k^3,\infty)\) d) \((g_k^{-1})'(k^3)=\frac{1}{6k^3}\)