Question

Find the type of monotonicity of $f\left(x\right)=\frac{x}{x^2-6x-16}$ when $xϵ[2,3].$

• A. decreasing
• B. increasing
• C. constant
• D. none

Answer 1

The correct option is A decreasing

$f\left(x\right)=\frac{x}{x^2-6x-16}$

$f^{\prime }\left(x\right)=\frac{(x^2-6x-16)-x(2x-6)}{(x^2-6x-16{)}^2}$

To find the interval where $f\left(x\right)$ increases,

$f^{\prime }\left(x\right)>0$
Or

$\frac{x^2-6x-16-x(2x-6)}{(x^2-6x-16{)}^2}>0$

Or

$x^2-6x-16-x(2x-6)>0$

$x^2-6x-16-2x^2+6x>0$

$-x^2-16>0$

Or

$x^2+16<0$

This is not possible as $x^2>0$ for all real values of $x$ .

Hence $f\left(x\right)$ is not increasing in any set of real values of $x$.

Therefore $f\left(x\right)$ is a strictly decreasing function.