Question

Consider the functions $f\left(x\right)=\{\begin{array}{cc}|x|,& x\le -1\\ x^{1/5}& -1<x\le 1\\ (2-x{)}^3,& x>1\end{array}\text{and}g\left(x\right)=\frac{1}{(x+2{)}^3}-2x-\cos x$. Let $p$ be the number of critical points on the graph of $f\left(x\right)\text{and}q$ be the number of solutions of $g\left(x\right)=0$, then which of the following option(s) is/are correct?

• A. $q=2$
• B. $p=2$
• C. $p=3$
• D. $q=1$

Answer 1

Answer: (A), (C)

$p=3$

$f\left(x\right)$ is discontinuous at $x=–1$, so not a critical point.
$x=1$ is a critical point being a sharp corner point
$x=0$ is a critical point, tangent being vertical
$x=2$ is a critical point, tangent being horizontal
$g^{\prime }\left(x\right)=-\frac{3}{(x+2{)}^4}-2+\sin x<0\forall xϵR-\{-2\}$
$g(-\infty )=\infty ,g\left(\infty \right)=-\infty ,g\left(x\right)$ is decreasing
Line $x=-2$ is the asymptote.
$\Rightarrow 2$ solutions