Help with question

f(x) = x^2    write f(x)  as y

y = x^2       get x by itself by taking the square root of both sides

√y =  x        "exchange"  x and y

√x = y         for y, write f^-1(x)

√x  =  f^-1(x)

In the original function, the domain is [-2, 2]  and the range is [0, 4]

In the inverse function,  the domain is  [0, 4]  and the range is [0, 2]

Here's a graph of both : https://www.desmos.com/calculator/rsveooxu3z

The first answer is the correct one  →  equals √x

BTW  -  good to see you.......how are you doing???

Hi Brittany, itr is REALLY good to see you again :))

Chris, I am not so sure about your answer.....

$If\;\;y=x^2\qquad then\qquad x=\pm\sqrt y$

If the domain of the original function was [0,2] then   $x=+\sqrt y$

And that would certainly have an inversese function.

BUT

given the original domain restriction of [-2,2]  the inverse becomes...

$y=\pm\sqrt x \qquad \mbox{and this is not a function at all}$

(It does not pass the vertical line test)

So I think, 

If f(x) = x² on the domain [-2,2] then f^-1 does not exist.

I could be wrong - I will certainly consider further discussion on the topic.