Question

$y=\frac{3x^2+4x+4}{x^2+x+1}$.Let $y_{max}=k,y_{min}=m$.Find $k+3\left(m\right)$?

Answer 1

$y=\frac{3x^2+4x+4}{x^2+x+1}=4-\frac{x^2}{x^2+x+1}$
$\frac{dy}{dx}=-\frac{(x^2+x+1)2x-x^2(2x+1)}{(x^2+x+1{)}^2}=-\frac{x(x+2)}{(x^2+x+1{)}^2}$
For max. or min. of $y$, $\frac{dy}{dx}=0$
$\Rightarrow x=0,2\Rightarrow y(x=0)=4,y(x=-2)=\frac{8}{3}$
$\therefore y_{max}=4,y_{min}=\frac{8}{3}$