Answer :
Answer:
The solution is [tex](f^{-1})'(a) = \frac{2}{\pi } [/tex]
Step-by-step explanation:
From the question we are told that
The function is [tex]f(x) = 4 + x^2 + tan [\frac{ \pi x}{2} ][/tex] , -1 < x < 1 a = 4
Here we are told find [tex](f^{-1}) (a)[/tex]
Let equate
[tex]f(x) = a[/tex]
So
[tex]4 + x^2 + tan[\frac{\pi x }{2} ] = 4[/tex]
[tex]x^2 + tan[\frac{\pi x }{2} ] = 0[/tex]
For the equation above to be valid x must be equal to 0
Now when x = 0
[tex]f(0) = 4+0^2 + tan [\frac{ \pi * 0}{2} ][/tex]
=> [tex]f(0) = 4[/tex]
=> [tex]0 = f^{-1} (4)[/tex]
Differentiating f(x)
[tex]f(x)' = 0 + 2x + sec^2 (\frac{\pi x}{2} )\cdot \frac{\pi}{2}[/tex]
Now
since [tex]0 = f^{-1} (4)[/tex]
We have
[tex]f(0)' = 0 + \frac{\pi }{2} sec^2 (0)[/tex]
[tex]f(0)' = \frac{\pi }{2} [/tex]
Now
[tex](f^{-1})'(a) = \frac{1}{(\frac{\pi}{2} )}[/tex]
[tex](f^{-1})'(a) = \frac{2}{\pi } [/tex]
The value of the function from the given function is [tex]f^{-1}'(a)=\frac{2}{\pi}[/tex]
Given the function [tex]f(x)=4+x^2+tan(\frac{\pi x}{2} )[/tex]
Given that f(x) = a = 4, hence the equation becomes;
[tex]4=4+x^2+tan(\frac{\pi x}{2} )\\x^2+tan(\frac{\pi x}{2})=0[/tex]
Since the equation must be true for x = 0, hence;
[tex]f(0)=4+0^2+tan(\frac{\pi (0)}{2} )\\f(0) = 4[/tex]
[tex]f^{-1}(4)=0[/tex]
Next is to differentiate the function with respect to x as shown:
[tex]f'(x)=0+2x+\frac{\pi}{2}sec^2(\frac{\pi x}{2} ) \\f'(0)=0+2(0)+\frac{\pi}{2}sec^2(\frac{\pi (0)}{2} ) \\f'(0)=\frac{\pi}{2}(1)\\f'(0)= \frac{\pi}{2}[/tex]
From the calculation above, we can see that;
[tex]f^{-1}'(a)=\frac{1}{\frac{\pi}{2} } \\f^{-1}'(a)=\frac{2}{\pi}[/tex]
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