Question

If $x$ is real, find the maximum and minimum values of $\frac{x^2+14x+9}{x^2+2x+3}.$

Answer 1

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

$\Rightarrow y(x^2+2x+3)=x^2+14x+9$

$\Rightarrow (y-1)x^2+2(y-7)x+3y-9=0$

Since, $x$ is real, so discriminant of above equation will be greater than or equal to 0.

$D\ge 0$

$\Rightarrow 4(y-7{)}^2-4(y-1\left)\right(3y-9)\ge 0$

$\Rightarrow (y-7{)}^2-(y-1\left)\right(3y-9)\ge 0$

$\Rightarrow y^2+49-14y-3(y^2-4y+3)\ge 0$

$\Rightarrow -2y^2-2y+40\ge 0$

$\Rightarrow y^2+y-20\le 0$

$\Rightarrow y^2+5y-4y-20\le 0$

$\Rightarrow y(y+5)-4(y+5)\le 0$

$\Rightarrow (y+5)(y-4)\le 0$

$\Rightarrow y\in [-5,4]$

So, the maximum value of $y$ is 4 and minimum value of $y$ is -4