Question

Let $h\left(x\right)=f\left(x\right)-{\left\{f\left(x\right)\right\}}^2+{\left\{f\left(x\right)\right\}}^3$ for all real values of $x$. Then

• A. $h$ is increasing whenever is $f\left(x\right)$ increasing
• B. $h$ is increasing whenever $f^{\prime }\left(x\right)<0$
• C. $h$ is decreasing whenever $f$ is decreasing
• D. nothing can be said in general

Answer 1

Answer: (A), (C)

The correct options are
A $h$ is increasing whenever is $f\left(x\right)$ increasing
C $h$ is decreasing whenever $f$ is decreasing

$h\left(x\right)=f\left(x\right)-{\left\{f\left(x\right)\right\}}^2+{\left\{f\left(x\right)\right\}}^3$
$\Rightarrow h^{\prime }\left(x\right)=f^{\prime }\left(x\right)-2f\left(x\right)f^{\prime }\left(x\right)+3\left[f\right(x\left){]}^2{f}^{\prime }\right(x)$
$\Rightarrow h^{\prime }\left(x\right)=f^{\prime }\left(x\right)\left[3\right\{f\left(x\right){\}}^2-2f\left(x\right)+1]$
Case 1. If $f^{\prime }\left(x\right)>0$ i.e. f is increasing
Then $h^{\prime }\left(x\right)>0$, since $\left[3\right\{f\left(x\right){\}}^2-2f\left(x\right)+1]>0[\because$ discriminant of this quadratic is negative $]$
Case 2. If $f^{\prime }\left(x\right)<0$ i.e. f is decreasing
Then $h^{\prime }\left(x\right)<0$, since $\left[3\right\{f\left(x\right){\}}^2-2f\left(x\right)+1>0]$