Question

If $f\left(x\right)=\frac{x^2-8}{x-3}$ is an increasing function, then

• A. $x\in (-\infty ,2]\cup [4,\infty )$
• B. $x\in (2,4)$
• C. $x\in (-\infty ,3)$
• D. $x\in (2,\infty )$

Answer 1

The correct option is A $x\in (-\infty ,2]\cup [4,\infty )$

$f\left(x\right)=\frac{x^2-8}{x-3}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(x-3)\left(2x\right)-(x^2-8)}{(x-3{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x^2-6x+8}{(x-3{)}^2}=\frac{(x-2)(x-4)}{(x-3{)}^2}$
For $f\left(x\right)$ to be increasing, $f^{\prime }\left(x\right)\ge 0$


$\Rightarrow x\in (-\infty ,2]\cup [4,\infty )$