Question

Let $g:R\to R$ be a differentiable function with $g\left(0\right)=0,g^{\prime }\left(0\right)=0\text{and}{g}^{\prime }\left(1\right)\ne 0.$ Let
$f\left(x\right)=\{\begin{array}{c}\frac{x}{|x|}g\left(x\right),x\ne 0\\ 0,x=0\end{array}$
and $h\left(x\right)=e^{|x|}$ for all $x\in R.$ Let $(f\circ h)\left(x\right)$ denotes $f\left(h\right(x\left)\right)$ and $(h\circ f)\left(x\right)$ denotes $h\left(f\right(x\left)\right).$ Then which of the following is (are) true?

• A. $f$ is differentiable at $x=0$
• B. $h$ is differentiable at $x=0$
• C. $f\circ h$ is differentiable at $x=0$
• D. $h\circ f$ is differentiable at $x=0$

Answer 1

Answer: (A), (D)

The correct options are
A $f$ is differentiable at $x=0$
D $h\circ f$ is differentiable at $x=0$

$g$ is differentiable, hence, continuous.
$\Rightarrow g\left(0\right)=g\left(0^{+}\right)=g\left(0^{-}\right)=0$

Option $\left(1\right)$:
$\text{LHD}=f^{\prime }\left(0^{-}\right)=\underset{\delta \to 0}{lim}\frac{f\left(0\right)-f(0-\delta )}{\delta }$
$=\underset{\delta \to 0}{lim}\frac{0-(-g(-\delta \left)\right)}{\delta }=0$

$\text{RHD}=f^{\prime }\left(0^{+}\right)=\underset{\delta \to 0}{lim}\frac{f(0+\delta )-f\left(0\right)}{\delta }$
$=\underset{\delta \to 0}{lim}\frac{g\left(\delta \right)}{\delta }=0$
$\therefore f$ is differentiable at $x=0$


Option $\left(2\right)$:
$h\left(x\right)=\{\begin{array}{c}{e}^{-x},x<0\\ e^x,x\ge 0\end{array}$

$h^{\prime }\left(x\right)=\{\begin{array}{c}-e^{-x},x<0\\ e^x,x\ge 0\end{array}$

$h^{\prime }\left(0^{-}\right)=-1\text{and}{h}^{\prime }\left(0^{+}\right)=1$
$\therefore h$ is not differentiable at $x=0$


$f\left(h\right(x\left)\right)=g\left(e^{|x|}\right),\forall x\in R$

Option $\left(3\right)$:
$\text{LHD}=f^{\prime }\left(h\right(0^{-}\left)\right)=\underset{\delta \to 0}{lim}\frac{f\left(h\right(0\left)\right)-f\left(h\right(0-\delta \left)\right)}{\delta }$
$=\underset{\delta \to 0}{lim}\frac{g\left(1\right)-g\left(e^{\delta }\right)}{\delta }=-g^{\prime }\left(1\right)$

$\text{RHD}=f^{\prime }\left(h\right(0^{+}\left)\right)=\underset{\delta \to 0}{lim}\frac{f\left(h\right(0+\delta \left)\right)-f\left(h\right(0\left)\right)}{\delta }$
$=\underset{\delta \to 0}{lim}\frac{g\left(e^{\delta }\right)-g\left(1\right)}{\delta }=g^{\prime }\left(1\right)$
$\therefore f\circ h$ is not differentiable at $x=0$


$h\left(f\right(x\left)\right)=\{\begin{array}{c}{e}^{|g\left(x\right)|},x\ne 0\\ 1,x=0\end{array}$

Option $\left(4\right)$:
$\text{LHD}=h^{\prime }\left(f\right(0^{-}\left)\right)=\underset{\delta \to 0}{lim}\frac{h\left(f\right(0\left)\right)-h\left(f\right(0-\delta \left)\right)}{\delta }$
$=\underset{\delta \to 0}{lim}\frac{1-1}{\delta }=\underset{\delta \to 0}{lim}\frac{0}{\delta }=0$

$\text{RHD}=h^{\prime }\left(f\right(0^{+}\left)\right)=\underset{\delta \to 0}{lim}\frac{h\left(f\right(0+\delta \left)\right)-h\left(f\right(0\left)\right)}{\delta }$
$=\underset{\delta \to 0}{lim}\frac{1-1}{\delta }=\underset{\delta \to 0}{lim}\frac{0}{\delta }=0$
$\therefore h\circ f$ is differentiable at $x=0$