# Homework Help: Simplifying Inverse Functions

1. Jul 6, 2011

### frozonecom

1. The problem statement, all variables and given/known data
$f(x) = 3x^2-6$

We are asked to solve for the inverse function of the above function.

2. Relevant equations

3. The attempt at a solution
$y=3x^2-6$
$x=3y^2-6$
$\frac{3y^2=x+6}{3}$
$y^2 = \frac{x+6}{3}$
$\sqrt{y^2}= \frac{\sqrt{x+6}}{3}$
$y= \frac{\sqrt{x+6}}{\sqrt{3}}$
$y= \frac{\sqrt{x+6}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$
$y= \frac{\sqrt{3x+18}}{3}$
$f^{-1}(x) = \frac{\sqrt{3x+18}}{3}$

however, my teacher marked me wrong. I don't know why. I might ask her tomorrow but it's our mastery test and she might not explain it to me. Anyone care to tell me what wrong I did? Is it the rationalization? Is it supposed to have a radical denominator? Thank!

2. Jul 6, 2011

### HallsofIvy

What was the exact wording of the question? Since $3x^2- 6$ is not one-to-one, strictly speaking it does not have an inverse.

If we restrict the original function to $f(x)= 3x^2- 6$ for $x\ge 0$ and not defined for x< 0, then
$$f^{-1}(x)= \sqrt{\frac{x+ 6}{3}}= \frac{\sqrt{x+ 6}}{\sqrt{3}}= \frac{\sqrt{3x+ 18}}{3}$$

If we restrict the original function to $f(x)= 3x^2- 6$ for $x\le 0[/tex] and not defined for x> 0, then $$f^{-1}(x)= -\sqrt{\frac{x+ 6}{3}}= -\frac{\sqrt{x+ 6}}{\sqrt{3}}= -\frac{\sqrt{3x+ 18}}{3}$$ Your teacher might accept $$f^{-1}(x)= \pm\frac{\sqrt{x+ 6}}{\sqrt{3}}= \pm\frac{\sqrt{3x+ 18}}{3}$$ Last edited by a moderator: Jul 6, 2011 3. Jul 6, 2011 ### frozonecom oooh. Her instructions was that we need to change a list of functions into inverse functions, nothing more. I think she was exercising us in changing functions to inverse functions but did not bother to tell us that functions need to be one-to-one to have an inverse. She didn't also provide any other information like restrictions [itex] x \leq 0$ or $x \geq 0$
so I guess she'll have to accept my answer. Thanks for helping! :)

4. Jul 7, 2011

### frozonecom

just wanna say that she accepted it. :) Thanks!