Re: inverse function problem

Quote:

Originally Posted by

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

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

Edited: Corrected.

Re: inverse function problem

Quote:

Originally Posted by

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

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.

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 $\displaystyle f(1/2)=7/8$ . I supposed (without computing) $\displaystyle f(7/8)=1/2$ .