1. Inverse Functions

So, I need to find (F^-1)'(a) of

f(x) = 3 x^2 + tan(πx/2) , -1< x < 1, a=3

I have worked the problem to this point

(f^-1)'(3) = 1/2x + sec^2(πx/2)π/2
= 1/2(0) + sec^2(π(0)/2)π/2

but my book and my professor tell me that the answer is 2/π, but I don't see how that can be. Shouldn't the answer be 1/0 because the 0's in the denominator give you 0 + 0, which is 0?

2. Originally Posted by zachb
So, I need to find (F^-1)'(a) of

f(x) = 3 x^2 + tan(πx/2) , -1< x < 1, a=3

I have worked the problem to this point

(f^-1)'(3) = 1/2x + sec^2(πx/2)π/2
= 1/2(0) + sec^2(π(0)/2)π/2

but my book and my professor tell me that the answer is 2/π, but I don't see how that can be. Shouldn't the answer be 1/0 because the 0's in the denominator give you 0 + 0, which is 0?
is f(x)=3*x^2 + tan(πx/2)

or should it be

f(x)=3+x^2 + tan(πx/2)

3. yes it should be

f(x)=3+x^2 + tan(πx/2)

I'm trying to find (f^-1)'(a)

so,

1/f^-1(f^-1(a))= 1/2x + sec^2(πx/2)π/2
1/2(0) + sec^2(π(0)/2)π/2

I just want to know why the answer is 2/π.

4. Originally Posted by zachb
yes it should be

f(x)=3+x^2 + tan(πx/2)

I'm trying to find (f^-1)'(a)

so,

1/f^-1(f^-1(a))= 1/2x + sec^2(πx/2)π/2
1/2(0) + sec^2(π(0)/2)π/2

I just want to know why the answer is 2/π.
f'(x)=2x+sec^2(nx/2)*(n/2)
f'(0)=2(0)+sec^2(n0/2)*(n/2)=0+1*(n/2)=n/2

f(0)=3 so f^{-1}(3)=0

(f^{-1})'(3)=1/(f'(f^{-1}(3)))=1/f'(0)=1/(n/2)=2/n

5. I guess what I should ask is why does n(0)/2 become n/2 instead of 0?

6. Originally Posted by zachb
I guess what I should ask is why does n(0)/2 become n/2 instead of 0?
it doesn't

n*0/2=0

sec^2(n*0/2)=sec^2(0)=1

so

[sec^2(n*0/2)]*(n/2)=[1]*(n/2)=n/2