<s>[INST] 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 !!!
 [/INST] 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 $1-\frac{12}{u}=1-\frac{12}{15/4}=-\frac{11}{5}$.
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