# Graph problem

1. Feb 5, 2008

### temaire

1. The problem statement, all variables and given/known data

2. Relevant equations
Not necessary

3. The attempt at a solution
The answers that I have up there are correct, but I was wondering why graph 2 did not look something like this:

2. Feb 5, 2008

### rock.freak667

Let y=x^2 so x=sqrt(y)

so $f^{-1}(x)=\sqrt{x}$ and $f^{-1}(f(x))=\sqrt{x^2}$

3. Feb 6, 2008

### Tedjn

In fact, the inverse of $y=x^2$ is $y=\sqrt{x}, x \geq 0$ and $y=-\sqrt{x}, x < 0$. It doesn't matter in the end, however, because the definition of an inverse says that $f^-1(f(x)) = x$

4. Feb 6, 2008

### HallsofIvy

Staff Emeritus
Actually, that's a seriously bad question! Since y= x2 is not one-to-one, it does not have an inverse! temaire, the graph you showed would be the "inverse function" to x2 but, as you can see from the graph, there are two values of x for each positive y and so it is not a function.,