Question

If $x$ is real, then the range of the function $f\left(x\right)=\frac{x^2+14x+9}{x^2+2x+3}$ is

• A. $(4,5)$
• B. $(-4,5)$
• C. $[-5,4]$
• D. None of the above

Answer 1

The correct option is C $[-5,4]$

Let $f\left(x\right)=y$
$\frac{x^2+14x+9}{x^2+2x+3}=y\Rightarrow x^2+14x+9=x^{2}y+2xy+3y(\because x^2+2x+3>0)\Rightarrow x^2(y-1)+2x(y-7)+3y-9=0$
For $x\in R$
$D\ge 0\Rightarrow 4(y-7{)}^2-4(3y-9\left)\right(y-1)\ge 0\Rightarrow 4(y^2+49-14y)-4(3y^2+9-12y)\ge 0\Rightarrow y^2-14y+49-(3y^3-12y+9)\ge 0\Rightarrow 2y^2+2y-40\le 0\Rightarrow (y+5\left)\right(y-4)\le 0\therefore y\in [-5,4]$

Hence, the range of $f\left(x\right)$ is $[-5,4]$