Question

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

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

Answer 1

Answer: (A), (D)

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

given g(0)=0 $g^{{}^{\prime }}\left(0\right)=0g^{{}^{\prime }}\left(1\right)\ne 1$
$f\left(x\right)=-g\left(x\right)x<0g\left(x\right)x>00x=0$
For option (A) R.H.D=L.H.D(by applying basic differntiation definition).
For option (B) $h\left(x\right)=-e^{-x}<0,e^{x}forx>0,0forx=0$
L.H.D is not equal to R.H.D in this case.
For option (C)
foh=f(h(u))=g(h(u)), $h\left(u\right)>0$
foh is not differentiable at x=0 (since L.H.D is not equal to R.H.D)
For option (D)$R.H.D{lim}_{t\to 0}\frac{h\left(f\right(t\left)\right)-h\left(f\right(0\left)\right)}{t}={lim}_{t\to 0}\frac{h\left(g\right(t\left)\right)-h\left(0\right)}{t}=0L.H.D{lim}_{t\to 0}\frac{h\left(f\right(-t\left)\right)-h\left(f\right(0\left)\right)}{-t}={lim}_{t\to 0}\frac{h(-g(-t\left)\right)-1}{-t}-0$
since L.H.D =R.H.D it is differentiable