Question

Find the maximum value of the function $\frac{x^2+14x+9}{x^2+2x+3}$ over $R$.

Answer 1

Let $y=\frac{x^2+14x+9}{x^2+2x+3}$

diff. w. r. to x

$\frac{dy}{dx}=\frac{\left(x^2+2x+3\right)\left(2x+19\right)-\left(x^2+14x+9\right)\left(2x+2\right)}{{\left(x^2+2x+3\right)}^2}$

$\frac{dy}{dx}=\frac{\left(2x^3+14x^2+4x^2+28x+6x+42\right)-\left(2x^3+2x^2+28x^2+28x+18x+18\right)}{{\left(x^2+2x+3\right)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{\left(2x^3+18x^2+34x+42\right)-\left(2x^3+30x^2+46x+18\right)}{{\left(x^2+2x+3\right)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{2x^3+18x^2+34x+42-2x^3-30x^2-46x-18}{{\left(x^2+2x+3\right)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{-12x^2-12x+24}{{\left(x^2+2x+3\right)}^2}\_\_\_\_\_\_\_\_\_\left(i\right)$

for max & min

$\frac{dy}{dx}=0.$

$\Rightarrow \frac{-12x^2-12x+24}{{\left(x^2+2x+3\right)}^2}=0$

$\Rightarrow -12x^2-12x+24=0$

$\begin{array}{c}\Rightarrow -12\left(x^2-x-2\right)=0\\ \Rightarrow x^2-x-2=0\\ \Rightarrow x^2-2x+x-2=0\\ \Rightarrow x\left(x-2\right)+1\left(x-2\right)=0\\ \Rightarrow \left(x+1\right)\left(x-2\right)=0\\ \Rightarrow x=-1,2\end{array}$

when x=-1

$y=\frac{{\left(-1\right)}^2+14\left(-1\right)+9}{{\left(-1\right)}^2+2\left(1\right)+3}$

$=\frac{1-19+9}{1-2+3}$

$=\frac{10-19}{4-2}$

$=\frac{-4}{2}$

=$-2$

when x=2

$\begin{array}{c}y=\frac{2^2+14\times 2+9}{2^2+2\times 2+3}\\ =\frac{4+28+9}{4+4+3}\\ =\frac{32+9}{11}\\ =\frac{41}{11}\\ y=\frac{41}{11}\end{array}$

Hence,

Max value

$y=\frac{41}{11}$