Applications of Derivatives
Monotonicity of Rational Function
nta_pyq_2024_jan
Grade 12
Question:
The function $f(x)=\dfrac{x}{x^2-6x-16}$, $x\in\mathbb{R}-\{-2,8\}$
decreases in $(-2,8)$ and increases in $(-\infty,-2)\cup(8,\infty)$
decreases in $(-\infty,-2)\cup(-2,8)\cup(8,\infty)$
decreases in $(-\infty,-2)$ and increases in $(8,\infty)$
increases in $(-\infty,-2)\cup(-2,8)\cup(8,\infty)$
Step-by-Step Solution
Key Concept: $f'(x)=\frac{(x^2-6x-16)-x(2x-6)}{(x^2-6x-16)^2}=\frac{-(x^2+16)}{(x^2-6x-16)^2}$. Since $x^2+16>0$ always, $f'(x)<0$ everywhere in the domain.
$f'(x)=\frac{-(x^2+16)}{(x^2-6x-16)^2}<0$ for all $x$ in the domain. So $f$ is decreasing on $(-\infty,-2)\cup(-2,8)\cup(8,\infty)$.
Correct Answer: 2