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AP Calculus AB Quiz
Practice Differentiating Inverse Functions in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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Given g(x)=x4+2 and g(1)=3, what is (g−1)′(3)?
This quiz focuses on Differentiating Inverse Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
Given g(x)=x4+2 and g(1)=3, what is (g−1)′(3)?
Explanation: 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.
Let f(x)=x3+1 and f(2)=9. What is (f−1)′(9)?
Explanation: 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)=3x2, so f'(2) = 12, thus (f−1)′(9)=121. 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.
Let f(x)=x3+2x−5 and f(2)=7. What is (f−1)′(7)?
Explanation: 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.
Let f(x)=x3+2x−5 and f(2)=7. If g=f−1, what is g′(7)?
Explanation: This problem requires finding the derivative of an inverse function using the inverse function derivative theorem. Since g = f^(-1) and f(2) = 7, we know that g(7) = 2. The key relationship is that g'(7) = 1/f'(g(7)) = 1/f'(2). To find f'(2), we differentiate f(x) = x³ + 2x - 5 to get f'(x) = 3x² + 2, so f'(2) = 3(4) + 2 = 14. Therefore, g'(7) = 1/14. A common error is computing f'(7) instead of f'(2), which would give 1/149 rather than the correct 1/14. Remember: for inverse function derivatives, always evaluate f' at the point that maps to your input, not at the input itself.
Suppose f(x)=sinx+x and f(0)=0. If g=f−1, what is g′(0)?
Explanation: This problem requires finding the derivative of an inverse function at a specific point. Given f(0) = 0 and g = f^(-1), we have g(0) = 0. The inverse derivative formula gives us g'(0) = 1/f'(g(0)) = 1/f'(0). To find f'(x) from f(x) = sin(x) + x, we differentiate to get f'(x) = cos(x) + 1, so f'(0) = cos(0) + 1 = 1 + 1 = 2. Thus g'(0) = 1/2. Students might mistakenly think the answer is 2 by computing cos(0) + 1 directly without taking the reciprocal. The strategy: always remember that inverse function derivatives involve taking the reciprocal of the original function's derivative.
Let f(x)=ln(x)+x with f(1)=1. If g=f−1, what is g′(1)?
Explanation: To find the derivative of the inverse function g at x = 1, we use the inverse derivative relationship. Given f(1) = 1 and g = f^(-1), we know g(1) = 1. The formula g'(1) = 1/f'(g(1)) = 1/f'(1) applies here. Differentiating f(x) = ln(x) + x yields f'(x) = 1/x + 1, so f'(1) = 1/1 + 1 = 2. Therefore, g'(1) = 1/2. A tempting error is to substitute into the expression 1 + 1/g(1), which would give 2 rather than 1/2. Always remember: the derivative of an inverse function equals the reciprocal of the original function's derivative, evaluated at the appropriate point.
Let f be differentiable and one-to-one with f(2)=5 and f′(2)=3; what is (f−1)′(5)?
Explanation: This problem requires finding the derivative of an inverse function at a specific point. The key relationship for inverse function derivatives is (f−1)′(b)=f′(a)1 where f(a)=b. Since we're given f(2)=5 and f′(2)=3, we need (f−1)′(5). Because f(2)=5, we know that f−1(5)=2, so we evaluate at a=2. Therefore, (f−1)′(5)=f′(2)1=31. Choice A (3) is tempting because it's f′(2), but the inverse derivative requires the reciprocal. Remember: to find (f−1)′ at a point, take the reciprocal of f′ at the corresponding input value.
A differentiable one-to-one function p satisfies p(0)=7 and p′(0)=5; compute (p−1)′(7).
Explanation: This problem involves finding the derivative of an inverse function at a specific value. The inverse function derivative formula tells us (p^{-1})'(b) = rac{1}{p'(a)} when p(a)=b. We have p(0)=7 and p′(0)=5, and need (p−1)′(7). Since p(0)=7, we get p−1(7)=0, so we use a=0. Thus (p^{-1})'(7) = rac{1}{p'(0)} = rac{1}{5}. Choice B (5) represents p′(0) without the reciprocal, a common error. To find an inverse derivative: locate the input that produces your output, then take the reciprocal of the derivative there.
If u is differentiable and one-to-one with u(8)=2 and u′(8)=0.1, what is (u−1)′(2)?
Explanation: To find the derivative of an inverse function, apply (u^{-1})'(b) = rac{1}{u'(a)} when u(a)=b. Given u(8)=2 and u′(8)=0.1, we seek (u−1)′(2). Since u(8)=2, we have u−1(2)=8, so we evaluate at a=8. Therefore, (u^{-1})'(2) = rac{1}{u'(8)} = rac{1}{0.1} = 10. Choice A (0.1) represents u′(8) itself rather than its reciprocal. The consistent pattern: inverse derivatives equal one divided by the original derivative at the pre-image point.
A one-to-one differentiable function g satisfies g(−1)=4 and g′(−1)=−2; find (g−1)′(4).
Explanation: This question asks for the derivative of an inverse function using the inverse function derivative formula. The fundamental relationship states that (g−1)′(b)=g′(a)1 when g(a)=b. We're told g(−1)=4 and g′(−1)=−2, and we need (g−1)′(4). Since g(−1)=4, we have g−1(4)=−1, so we use a=−1 in our formula. Thus (g−1)′(4)=g′(−1)1=−21=−21. Choice A (−2) incorrectly uses g′(−1) directly without taking the reciprocal. The strategy: identify which original input gives your desired output, then take the reciprocal of the derivative at that input.
Let q(x)=x+3 and q(6)=3. What is (q−1)′(3)?
Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by q′(b)1, where q(b)=a. Here, q(6)=3, so b=6 and a=3. Compute q′(x)=2x+31, so q′(6)=61, thus (q−1)′(3)=6. A tempting distractor is B, 61, 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.
Given f(x)=x5−x and f(1)=0, what is (f−1)′(0)?
Explanation: 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) = 0, so b = 1 and a = 0. Compute f′(x)=5x4−1, so f′(1)=4, thus (f−1)′(0)=41. 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 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.
Let h(x)=x2+2x (restricted to x≥−1) and h(3)=15. What is (h−1)′(15)?
Explanation: 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 h(b) = a. Here, h(3) = 15, so b = 3 and a = 15. Compute h'(x) = 2x + 2, so h'(3) = 8, thus (h⁻¹)'(15) = 1/8. A tempting distractor is A, 8, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to h, not to h⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.
Let t be differentiable and one-to-one with t(−4)=−6 and t′(−4)=π; what is (t−1)′(−6)?
Explanation: Finding this inverse derivative uses the formula (t^{-1})'(b) = rac{1}{t'(a)} where t(a)=b. We have t(−4)=−6 and t′(−4)=pi, and need (t−1)′(−6). Because t(−4)=−6, we get t−1(−6)=−4, so we use a=−4. Thus (t^{-1})'(-6) = rac{1}{t'(-4)} = rac{1}{pi}. Choice A (pi) is the original derivative value without reciprocation, a frequent mistake. Remember: inverse function derivatives always involve taking the reciprocal of the original derivative at the corresponding point.
Given f(x)=ex+x and f(0)=1, let g=f−1. What is g′(1)?
Explanation: This problem involves finding the derivative of an inverse function using the fundamental relationship between f and its inverse g. Since f(0) = 1 and g = f^(-1), we have g(1) = 0. The inverse derivative formula states g'(1) = 1/f'(g(1)) = 1/f'(0). Differentiating f(x) = e^x + x gives f'(x) = e^x + 1, so f'(0) = e^0 + 1 = 1 + 1 = 2. Thus g'(1) = 1/2. Students might incorrectly compute e^0 + 0 = 1, forgetting that the derivative of x is 1, not 0. The key insight: always differentiate the original function completely before evaluating at the required point.
Let f(x)=x3−4x and f(2)=0. If g=f−1, what is g′(0)?
Explanation: 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.
For f(x)=x5−x, f(1)=0. If g=f−1, what is g′(0)?
Explanation: This problem asks for the derivative of an inverse function at a specific point. Since g = f^(-1) and f(1) = 0, we have g(0) = 1. The inverse function derivative formula tells us that g'(0) = 1/f'(g(0)) = 1/f'(1). Finding f'(x) from f(x) = x⁵ - x gives f'(x) = 5x⁴ - 1, so f'(1) = 5(1) - 1 = 4. Thus g'(0) = 1/4. Students often mistakenly compute f'(0) = -1 and get -1 as the answer, forgetting to evaluate f' at g(0) = 1, not at 0. The strategy is: find where the inverse function maps your input, then evaluate f' at that point.
Given r(x)=ln(x+1) and r(e−1)=1, find (r−1)′(1).
Explanation: 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.
If t(x)=x3+3x and t(1)=4, what is (t−1)′(4)?
Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by derivative of the original function at b1, where t(b)=a. Here, t(1)=4, so b=1 and a=4. Compute t′(x)=3x2+3, so t′(1)=6, thus (t−1)′(4)=61. 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.
Let f(x)=x+x1 for x>0 and f(1)=2. Find (f−1)′(2).
Explanation: 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.