This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where f(b) = a. Here, f(2) = 9, so b = 2 and a = 9. Compute $f'(x) = 3x^2$, so f'(2) = 12, thus $(f^{-1})'(9) = \frac{1}{12}$. A tempting distractor is A, 12, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to f, not to $f^{-1}$. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where g(b) = a. Here, g(1) = 3, so b = 1 and a = 3. Compute g'(x) = 4x³, so g'(1) = 4, thus (g⁻¹)'(3) = 1/4. A tempting distractor is A, 4, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to g, not to g⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point $a$ is given by $ \frac{1}{\text{derivative of the original function at } b} $, where $ t(b) = a $. Here, $ t(1) = 4 $, so $ b = 1 $ and $ a = 4 $. Compute $ t'(x) = 3x^2 + 3 $, so $ t'(1) = 6 $, thus $ (t^{-1})'(4) = \frac{1}{6} $. A tempting distractor is A, $ 6 $, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to $ t $, not to $ t^{-1} $. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where f(b) = a. Here, f(1) = 2, so b = 1 and a = 2. Compute f'(x) = 1 - 1/x², so f'(1) = 0, thus (f⁻¹)'(2) = 1/0, which is undefined. A tempting distractor is A, 0, which might be chosen if one mistakenly takes the reciprocal of 0 as 0, but actually it indicates the inverse is not differentiable there. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where $r(b) = a$. Here, $r(e-1) = 1$, so b = e-1 and a = 1. Compute $r'(x) = 1/(x+1)$, so $r'(e-1) = 1/e$, thus $(r^{-1})'(1) = e$. A tempting distractor is A, $1/e$, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to r, not to r⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

To find the derivative of the inverse function g at x = 0, we use the inverse derivative relationship. Since f(2) = 0 and g = f^(-1), we know g(0) = 2. The formula g'(0) = 1/f'(g(0)) = 1/f'(2) applies. Differentiating f(x) = x³ - 4x gives f'(x) = 3x² - 4, so f'(2) = 3(4) - 4 = 12 - 4 = 8. Therefore, g'(0) = 1/8. A common error is computing f'(0) = -4 and getting -1/4, but we must evaluate f' at the point g(0) = 2. Remember: the input to g' determines where we evaluate f', not the input value itself.

This question tests understanding of inverse function differentiation. The key formula is $(v^{-1})'(b) = \frac{1}{v'(a)}$ where $v(a) = b$. Since $v(8) = -3$, we know $v^{-1}(-3) = 8$, and we need $(v^{-1})'(-3)$. Applying the formula with $a = 8$ and $b = -3$: $(v^{-1})'(-3) = \frac{1}{v'(8)} = \frac{1}{5/9} = \frac{9}{5}$. Choice B ($\frac{5}{9}$) is $v'(8)$ without reciprocation, a frequent error. Remember: inverse derivatives require reciprocating the original function's derivative at the pre-image point.

This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where g(b) = a. Here, g(0) = 1, so b = 0 and a = 1. Compute g'(x) = eˣ + 1, so g'(0) = 2, thus (g⁻¹)'(1) = 1/2. A tempting distractor is A, 2, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to g, not to g⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where f(b) = a. Here, f(2) = 7, so b = 2 and a = 7. Compute f'(x) = 3x² + 2, so f'(2) = 14, thus (f⁻¹)'(7) = 1/14. A tempting distractor is B, 14, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to f, not to f⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where q(b) = a. Here, q(1) = 2, so b = 1 and a = 2. Compute q'(x) = 3x² + 2x, so q'(1) = 5, thus (q⁻¹)'(2) = 1/5. A tempting distractor is A, 5, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to q, not to q⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.