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 !!!
 A: You do not need calculus/derivatives for this. Let $u=x^2+3x+6=\left(x+\frac32\right)^2+\frac{15}{4}$, and note that you would like to minimize $\frac{u-12}{u}=1-\frac{12}{u}$.
$u$ can take values in $[15/4,\infty)$. And $1-\frac{12}{u}$ is an increasing function for positive $u$.
It follows that the minimum occurs when $u=\frac{15}{4}$, which happens when $x=-\frac{3}{2}$. And the minimum value obtained is $1-\frac{12}{u}=1-\frac{12}{15/4}=-\frac{11}{5}$.
A: 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 $x>-\frac32$.
A: Another didactic way, we prove that $-11/5 \le y=f(x)<1, x \in \Re$.
Let $y=\frac{x^2+3x-6}{x^2+3x+6}$ and check that $x^2+3x+6 >0$
Then we can write $(y-1)x^2+3(y-1)x+6(y+1)=0, x \in \Re$
$\implies B^2 \ge 4AC \implies 9(y-1)^2\ge 24(y+1)(y-1).$
Case 1: $y<1$ we get $9y-9 \le 24y+24 \implies  y \ge -11/5$
and hence $-11/5 \le y <1$.
Case 2: when $y>1$ gives $y\le -11/5$ a Null.
