Question

If $x$ be real, find the maximum value of $\frac{x+2}{2x^2+3x+6}$.

Answer 1

Consider the given function.

$f\left(x\right)=\frac{x+2}{2x^2+3x+6}$

Clearly, the function $(2x^2+3x+6)$ is a quadratic function with $a=2>0$ and so it will have its minimum value at $x=-\frac{b}{2a}=-\frac{3}{4}$ and the minimum value will be,

$2{(-\frac{3}{4})}^2+3(-\frac{3}{4})+6=\frac{9}{8}-\frac{9}{4}+6=\frac{9-18+48}{8}=\frac{39}{8}$

We can see that,

$D<0$ and $2x^2+3x+6>0\forall x$

Hence,

$0<\frac{1}{(2x^2+3x+6)}<\frac{8}{39}\forall x\in R$

Therefore, $f\left(x\right)$ will have its maximum value at $x=-\frac{3}{4}$ and its value will be,

$\frac{(-\frac{3}{4})+2}{\left(\frac{39}{8}\right)}=\frac{5\times 8}{39\times 4}=\frac{10}{39}$

Hence, this is the required result.