Applications of Differential Calculus (KUMON Level O) I don't understand what I need to do to solve this problem:


  Show that equation $$\frac{x^2 - 3}{x-1}=0$$ has $2$ real solutions in $[-2,2]$.


I've found the derivative which is:

$$f'(x) = \frac{x^2 - 2x + 3}{(x-1)^2} &gt; 0$$

The function is discontinuous at $x=1$. 

I don't know what to do next.
Just solve 
$$\frac{x^2-3}{x-1}=0,$$
there is no need for calculus.
$Y’$ is alway greater than $0$ in the domain thus the function is always increasing in the domain at approaching $1$ from the negative side it is equal to infinity for the positive side it is equal to negative infinity and $f(-2)$ is smaller then $0$ so that is one real solution because the function is increasing  so it goes from some thing negative to infinity it will pass by $0$ and from negative infinity to posit if something it will go by $0$ too. thus there is two real solutions