How to obtain the inverse of a function? I am trying to solve this 
$$f(x)= \frac{-5}  {x^2 + 1}$$
The solution is 
$f^{-1} = \pm\frac{\sqrt{-x-5}} {\sqrt{x}}$
I have done 
$$x = \frac{-5}  {y^2 + 1}$$
$${(x)(y^2 +1)} = {-5} $$
$${y^2 +1} = {\frac{-5}{x}} $$
$${y^2} = {\frac{-5}{x} - 1} $$
$${y} = \sqrt{\frac{-5}{x} - 1} $$
But I do not get the final answer as yo can see, I need to use the inverse to get the range of the original function,can someone please guide me in how to solve this exercise
 A: You mentioned you want to find the range of the original function. To do this, recall that a fraction gets larger in absolute value as the denominator gets smaller. But the function always takes on negative values, so its minimum (if it has one) will occur when the denominator is as small as possible. $x^2+1$ is minimized at $x=0$, which gives the minimum as
$$\frac{-5}{0^2+1}=-5$$
There actually is no maximum value for the function, because the denominator can get arbitrarily large, and the function will get arbitrarily close to $0$ from the negative side, but never actually reach it. So the range, in interval notation, is $$[-5,0)$$

As for the inverse you found, it doesn't match the final answer because you forgot to consider the negative root for $y$ in
$$y^2 = \frac{-5}{x}-1$$
The two roots are
$$y = \pm \sqrt{\frac{-5}{x}-1}$$
This is equivalent to the given answer:
$$y = \pm \sqrt{\frac{-5}{x}-1} = \pm \sqrt{\frac{-5}{x}+\frac{-x}{x}} = \pm \sqrt{\frac{-x-5}{x}}$$$$$$
You can use the inverse function to get the same result for the range of the original function, using two ideas: if you're working in real numbers, you can't divide by zero, and you can't take the square root of a negative number. 
