Question

Find the greatest value of $\frac{x+2}{2x^2+3x+6}$ for real values of $x$.

Answer 1

Let $y=\frac{x+2}{2x^2+3x+6}$

$=\frac{dy}{dx}=\frac{2x^2+3x+6-(x+2)(4x+3)}{(2x^2+3x+6{)}^2}$

$=\frac{dy}{dx}=\frac{-2x^2-8x}{(2x^2+3x+6{)}^2}$

Now, for maxima, $\frac{dy}{dx}=0$

$\Rightarrow x=0,-2$

At $x=0,y=\frac{1}{3}$

At $x=-2,y=0$

Hence, the greatest value of $\frac{x+2}{2x^2+3x+6}=\frac{1}{3}$