# Find the domain of the inverse of a function

Find the domain of the inverse of the following function. The function is defined for x<=0

I found the inverse of the function to be:

for the inverse to exist the values inside the square root have to be positive, which happens if the denominator and numerator are both positive or both negative.

Therefore, when both are positive: -9x-4 > 0 and x-1 > 0

when both are negative: -9x-4 < 0 and x-1 < 0

Hence, I found the domain to be x>1 and x<-4/9.

However, the solution states that the is -4/9 < x < 1 . How is this the answer?

• It's impossible for both the numerator and denominator to be positive simultaneously. If they're both negative then you should get$$-9x-4\lt0\iff x\gt-\frac49$$$$x-1\lt0\iff x\lt1$$ – Peter Foreman Jun 20 '20 at 8:40

The domain of the inverse equals the range of the original function. Consider that$$f(x)=\frac{x^2-4}{x^2+9}=1-\frac{13}{x^2+9}\ge 1-\frac{13}9=-\frac49$$then also we have$$\frac{1}{x^2+9}\gt0\implies f(x)\lt1$$so the domain of $$f^{-1}$$ is $$-\frac49\le x\lt1$$
• I have effectively just applied the range of the function $x^2$ (which is $[0,\infty)$) to get the result above without a sketch. – Peter Foreman Jul 15 '20 at 12:37
Here is an alternative way to do so.Note that the domin of inverse function is the values that the function itself can produce.Also the function $$\frac{y-4}{y+9}$$ is strictly increasing for $$y>-9$$ so the minimum occures for $$x^2=0$$ and maximum occures when $$x$$ tends to infinity also the function is continues so it can produce anything in the interval $$[-\frac{4}{9},1)$$.