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(2x), 0 ≤ x ≤ /2, a = 1

Answer 1

The answer is:

`(f^(-1))^(\prime)(1)=DNE`

Step-by-step explanation:

We need to use the theorem for the inverse of the derivative:

`(f^(-1))^(\prime)(x)=(1)/(f^(\prime)(f^(-1)(x)))`

Then, we need to calculate the derivative of f and its inverse.

To find the derivative, we know that the derivative of cosine is negative sine, and by the chain rule:

`f^(\prime)(x)=-2\sin(2x)`

ANd to find the inverse, we call f(x) = y and solve for x:

`\begin{gathered} y=\cos(2x) \\ \cos^(-1)(y)=2x \\ \end{gathered}`
`x=(\cos^(-1)(y))/(2)`

Now, we switch the variables, and we have the inverse:

`f^(-1)(x)=(\cos^(-1)(x))/(2)`

And now, by the theorem:

`(f^(-1))^(\prime)(x)=(1)/(f^(\prime)(f^(-1)(x)))`

To find the derivative of the inverse in x = 1, we need to find first:

`f^(-1)(1)=(\cos^(-1)(1))/(2)=(0)/(2)=0`

Now:

`(f^(-1))^(\prime)(x)=(1)/(f^(\prime)(0))`
`f^(\prime)(0)=-2\sin(2\cdot0)=-2\cdot0=0`

And finally:

`(f^(-1))^(\prime)(x)=(1)/(0)=DNE`

Thus, the correct answer is DNE