Question

Find (f −1)'(a). f(x) = 4 + x2 + tan(πx/2), −1 < x < 1, a = 4

Answer 1

Answer:

The  solution  is   $(f^{-1})'(a) = \frac{2}{\pi }$

Step-by-step explanation:

From the question we are told that

      The  function is  $f(x) = 4 + x^2 + tan [\frac{ \pi x}{2} ]$ ,     -1 <  x  <  1  a =  4

Here we are told find  $(f^{-1}) (a)$

Let equate

     $f(x) = a$

So  

      $4 + x^2 + tan[\frac{\pi x }{2} ] = 4$

       $x^2 + tan[\frac{\pi x }{2} ] = 0$

For the equation above to be valid x must be equal to 0

  Now when x = 0

       $f(0) = 4+0^2 + tan [\frac{ \pi * 0}{2} ]$

=>      $f(0) = 4$

=>  $0 = f^{-1} (4)$

Differentiating  f(x)

     $f(x)' = 0 + 2x + sec^2 (\frac{\pi x}{2} )\cdot \frac{\pi}{2}$

Now

    since $0 = f^{-1} (4)$

We have

      $f(0)' = 0 + \frac{\pi }{2} sec^2 (0)$

       $f(0)' = \frac{\pi }{2}$

Now  

        $(f^{-1})'(a) = \frac{1}{(\frac{\pi}{2} )}$

        $(f^{-1})'(a) = \frac{2}{\pi }$