# Homework Help: Inverse of radical function

1. Jul 17, 2011

### Nelo

1. The problem statement, all variables and given/known data
Find the inverse of each of the following functions

2. Relevant equations

y= [sqrt] x^2 + 9
3. The attempt at a solution

y = [sqrt] x^2 +9
x= [sqrt] y^2 +9
x-3= y

I did the sqrt of 9, and sqrted y and its wrong.

The answer is apparently y=+/(plusminus) [sqrt]x^2 -9 .. How do you solve this?

2. Jul 17, 2011

### Nelo

anyone?

3. Jul 17, 2011

### tiny-tim

Hi Nelo!

(have a square-root: √ and try using the X2 icon just above the Reply box )
How did you get that second line?

4. Jul 17, 2011

### Nelo

Its an inverse...

5. Jul 17, 2011

### tiny-tim

No it isn't.

Write it out in full before trying to invert it.

6. Jul 17, 2011

### eumyang

Need some parentheses. I don't know if you mean
$y = \sqrt{x^2} + 9$
or
$y = \sqrt{x^2 + 9}$

Looks like the 9 is INSIDE the square root. You got it completely wrong from line 2 to line 3. From line 2, square both sides, and then subtract the 9. Then solve for y.

7. Jul 18, 2011

### HallsofIvy

There are two commonly taught ways to find inverses of functions:
1) First solve y= f(x) for x, then swap x and y.
2) First swap x and, y, then solve for y.

So given $f(x)= \sqrt{x^2+ 9}$, you can write that, first, as $y= \sqrt{x^2+ 9}$ then swap x and y to write $x= \sqrt{y^2+ 9}$.

However, it is not true that $\sqrt{a^2+ b^2}= a+ b$. For example, $\sqrt{9+ 16}= \sqrt{25}= 5$, not $\sqrt{9}+ \sqrt{16}= 3+ 4= 7$.

In order to solve $x= \sqrt{y^2+ 9}$ for y, start by squaring both sides.