Question

Use the function f and the given real number a to find (f -1)'(a). (Hint: See Example 5. If an answer does not exist, enter DNE.)f(x) = cos(3x), 0 ≤ x ≤ /3, a = 1

Answer 1

Given:

$f(x)=\cos(3x),0\leq x\leq\frac{\pi}{3},a=1$

Required: Derivative of inverse of x at the point x = 1

Explanation:

Use the formula

$(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}$

Substitute 1 for x.

$(f^{-1})^{\prime}(1)=\frac{1}{f^{\prime}(f^{-1}(1))}\text{ ...\lparen1\rparen}$

First, find the inverse of f(x).

Let y = f(x). Then y = cos(3x).

Exchange x and y gives x = cos(3y).

Solve for y, which will be the inverse of f(x).

$\begin{gathered} 3y=\cos^{-1}x \\ y=\frac{1}{3}\cos^{-1}x \end{gathered}$

So,

$f^{-1}(x)=\frac{1}{3}\cos^{-1}(x)$

Substitute 1 for x.

$\begin{gathered} f^{-1}(1)=\frac{1}{3}\cos^{-1}(1) \\ =0 \end{gathered}$

Now, find the derivative of f(x).

$f^{\prime}(x)=-3\sin(3x)$

Thus, from equation (1),

$\begin{gathered} (f^{-1})^{\prime}(1)=\frac{1}{f^{\prime}(f^{-1}(1))}\text{ ...\lparen1\rparen} \\ =\frac{1}{f^{\prime}(0)} \\ =\frac{1}{-3\sin(0)} \end{gathered}$

which is not defined. So, the answer not exists.