# inverse function problem

• December 16th 2011, 04:36 AM
angypangy
inverse function problem
I have a function

$f(x) = 4x^2 - x^3$

If the domain of this function is restricted to 0 < x < 8/3 then f-1(x) can be defined - ie inverse function.

I am told to find the gradient of $f^-1(x)$ at the point ( $\frac{7}{8}, \frac{1}{2}$)

My first thought was to work out the inverse function. But this looks hard? Is this what I should do? Or is there another way? I know that an inverse function is a reflection of function on y=x - can I somehow use this to calculate? If so how?

Angus
• December 16th 2011, 04:42 AM
FernandoRevilla
Re: inverse function problem
Quote:

Originally Posted by angypangy
I am told to find the gradient of $f^-1(x)$ at the point ( $\frac{7}{8}, \frac{1}{2}$)

$(f^{-1})'(7/8)=\frac{1}{f'(1/2)}=\ldots$

Edited: Corrected.
• December 16th 2011, 04:53 AM
angypangy
Re: inverse function problem
Quote:

Originally Posted by FernandoRevilla
$(f^{-1})'(1/2)=\frac{1}{f'(7/8)}=\ldots$

I have already calculated dy/dx of f(x) which is 8x - 3x^2. If I use the y value of f(x) when x=7/8, then I get:

8(1/2) - 3(1/2)^2 = 4 - 3/4 = 13/4.

The reciprocal of this is 4/13 - which is correct answer.
• December 16th 2011, 05:12 AM
FernandoRevilla
Re: inverse function problem
Quote:

Originally Posted by angypangy
I have already calculated dy/dx of f(x) which is 8x - 3x^2. If I use the y value of f(x) when x=7/8, then I get:

8(1/2) - 3(1/2)^2 = 4 - 3/4 = 13/4.

The reciprocal of this is 4/13 - which is correct answer.

You are right, I didn't check that $f(1/2)=7/8$ . I supposed (without computing) $f(7/8)=1/2$ .