Question

Find the interval of monotonicity of $f\left(x\right)=\frac{x}{x^2-6x-16}$

• A. $x\in (2,8)$
• B. $x\in (0,8)$
• C. $x\in (-2,8)$
• D. $x\in (-8,-2)$

Answer 1

The correct option is A $x\in (2,8)$

Given $y=\frac{x}{x^2-6x-16}$
It is not defined at points given by $x^2-6x-16=0$ or $(x+2)(x-8)=0$
$\therefore x=2,8$are to be excluded from the domain
Now $\frac{dy}{dx}=-\frac{x^2+16}{{(x+2)}^2{(x-8)}^2}=-i$ve for all real values of x except $-2,8$ .
Hence the function is decreasing in $R-\{2,8\}$