Inverse of radical function

[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?

[Nelo]

anyone?

[tiny-tim]

Hi Nelo! :smile:

(have a square-root: √ and try using the X2 icon just above the Reply box :wink:)
y = [sqrt] x^2 +9
x= [sqrt] y^2 +9

How did you get that second line? :confused:

[Nelo]

Its an inverse...

[tiny-tim]

No it isn't. :redface:

Write it out in full before trying to invert it. :smile:

[eumyang]

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



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.
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.

[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.