# Thread: Find the Inverse Function

1. ## Find the Inverse Function

Hello,

The function $y=f(x)$ is defined by $y=x^2 + 4$ for $-2< x < 0.$
Find the inverse function $f^-1(x)$ stating its range and domain, and sketch the inverse function $f^-1(x)$.

Now I do now really want the sketch since I will try finding that out myself.
But how do we find $f^-1(x)$ ?? What I remember is that im supposed to re-arrange the equation so that it is $x=$ .. but wont that change the domain and range ?

Kind Regards,

Z

2. The range and the domain swap for the inverse function.

Thus if you find the range of the original function it will be the domain of the inverse.

3. Domain of f^-1 is the range of f

Range of f^-1 is the domain of f

is the range of f^-1 this will be important as

x = + sqrt(y-4)

x= - sqrt(y-4)

f^-1(x) = -sqrt( x-4) Domain 4 < x < 8

Note the range is then -2 < y < 0

4. Thanks for the reply guys, thank you Calculus26 for the explanation.

Can you please tell me why you chose the negative sign at x= - sqrt(y-4) ??

Thanks

5. is the range of f^-1 this will be important as

x = + sqrt(y-4)
the range is negative

6. Oh so the graph will be a negative .. of course ..

I will solve the question now and post the sketch

Thanks

7. Ok I managed to get it all except for the range, why is y<8 ??
It is greater than 4 because Square root of 0 is undefined, so it must be greater than 4 ..

But any number greater than 8 will still give an answer ?

8. Consider the original problem

The domain was restricted to -2 < x < 0

Which gave a restricted range 4 < y < 8

These restrictions carry over to the inverse with the domains and ranges
reversed.

9. Oh yeahhhh

How can I forget such thing, thank you very very much !!

10. As a check the inverse of a function will be it's reflection in the line y=x

11. Yes I realized that when I sketched the function
Ill add that as a note for future reference, thank you.