Home / AP Calculus BC : 3.3 Differentiating Inverse Functions- Exam Style questions with Answer- MCQ

AP Calculus BC : 3.3 Differentiating Inverse Functions- Exam Style questions with Answer- MCQ

Question

                       

The graph of the increasing differentiable function f is shown above. Also shown is the line tangent to the graph of f at the point (2,4). Let g be the inverse of f. Which of the following statements about g′ is true?
A \({g}'(2)=\frac{2}{3}\)

B \({g}'(2)=\frac{3}{2}\)

C \({g}'(4)=\frac{2}{3}\)

D \({g}'(4)=\frac{3}{2}\)

Answer/Explanation

Ans:C

 

Question

Let f be the decreasing function defined by \(f(x)=−x^3−6x^2−12x+8\), where \(f(4)=−8\). If g is the inverse function of f, which of the following is a correct expression for \({g}'(−8)\) ?

A \({g}'(−8)=\frac{1}{{f}'(−8)}\)

B \({g}'(−8)=\frac{1}{{f}'(4)}\)

C \({g}'(−8)={f}'(4)\)

D \({g}'(−8)={f}'(−8)\)

Answer/Explanation

Ans:B

 This can be confirmed using the chain rule and the definition of an inverse function. Since \(f(g(x))=x\)

, it follows that \(\frac{d}{dx}f(g(x))={f}'(g(x)){g}'(x)=\frac{d}{dx}(x)=1⇒{g}'(x)=\frac{1}{{f}'(g(x))}\). Therefore, \({g}'(−8)=\frac{1}{{f}'(g(−8))}=\frac{1}{{f}'(4)}\).

Question

                       

The table above gives selected values for a differentiable and increasing function f and its derivative. If \(g(x)=f^{−1}(x)\) for all x, which of the following is a correct expression for \({g}'(0)\) ?

A \({g}'(0)={f}'(0)=2\)

B \({g}'(0)=\frac{1}{{f}'(0)}=\frac{1}{2}\)

C \({g}'(0)=\frac{1}{{f}'(−4)}=1\)

D \({g}'(0)=−\frac{{f}'(0)}{(f(0))^2}=−\frac{2}{9}\)

Answer/Explanation

Ans:C

This value can be confirmed using the chain rule and the definition of an inverse function. Since \(f(g(x))=x\)

Question

 If \(f(x)=\frac{\sqrt{x}}{2}\) and  \(g(x)=\cos x\),find \({[f(g(x))]}’\)
(A) \(\frac{\sin x}{4\sqrt{\cos x}}\)
(B) \(-\frac{\sin x}{4\sqrt{\cos x}}\)
(C) \(-\frac{\sin x}{2\sqrt{\cos x}}\)
(D) \(\frac{\sin x\sqrt{\cos x}}{4}\)

Answer/Explanation

Ans:(B)

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