What is the minimum value of the function $f(x)= \frac{x^2+3x-6}{x^2+3x+6}$? I was trying to use the differentiation method to find the minimum value of the person but it did not give any result, I mean when I differentiated this function and equated to zero for finding the value of $x$ when the function value would be minimum, it gave me an absurd relation like $-6=6$. 
How can we solve this? Why is the differentiation method not working here? Please help !!!
$f(x)= \frac{x^2+3x-6}{x^2+3x+6}$
$f'(x)= \frac{(x^2+3x+6)(2x+3)-(x^2+3x-6)(2x+3)}{(x^2+3x+6)^2}$
Then equated this to $0$ to find the value of $x$.
Thanks in advance !!!

Notice that we have $2x+3=0$ and you should check that $x=-\frac32$ is a minimum.
$$f'(x) = \frac{12(2x+3)}{(x^2+3x+6)^2}$$ shows that $x=-\frac32$ is the global minimum as the gradient is negative for $x<-\frac32$ and then positive for $x>-\frac32$.
