Discussion Forum

 Let  $g:R\rightarrow R$  be a differentiable  function with g(0)=0, g'(0)=0 and g'(1)≠ 0,$f(x)= \begin{cases}\frac{x}{|x|}g(x) & x \neq 0\\0 & x = 0\end{cases}$  and $h(x)=e^{|x|}$   For all x ε R. Let (foh)(x) denotes  f{h(x)}  and (hof)(x) denotes h{f(x)} . Then which of the following is/are true?

Answer:

Explanation:

 Here  ,   $f(x)=\begin{cases}g(x) ,& x > 0\\0, & x = 0\\-g(x),   & x<0\end{cases}$

              $f'(x)=\begin{cases}g'(x) ,& x \geq 0\\-g(x), & x < 0\end{cases}$

  $\therefore$     Option (a) is correct

 (b)           $h(x)=e^{|x|}=\begin{cases}e^{x} ,& x \geq 0\\e^{-x}, & x < 0\end{cases}$

   $\Rightarrow$      $h'(x)=\begin{cases}e^{x} ,& x \geq 0\\-e^{-x}, & x < 0\end{cases}$

 $\Rightarrow$     h' (0⁺ )=1

 and     h'(o^-)=-1

  So, h(x)  is not differentiable at x=0,

 $\therefore$   Option (b) is not correct.

  (c)    $(foh)(x)=f\left\{h(x)\right\},$  as     $h(x)>0$

                        $= \begin{cases}g(e^{x} ),& x \geq 0\\g(e^{-x}), & x < 0\end{cases}$

 $\Rightarrow$     $(foh)'(x)= \begin{cases}e^{x}g^{'}(e^{x}), & x \geq 0\\-e^{-x}g^{'}(e^{-x}) ,& x < 0\end{cases}$

$\Rightarrow$          $(foh)'(0^{+})=g'(1),(foh)'(0^{-})=-g'(1)$

             So, (foh)(x)  is not differentiable at x=0

   $\therefore$    Option (c) is not correct.

  (d)    $(hof)(x)=e^{|f(x)| }=\begin{cases}e^{|g(x)|}, & x \neq 0\\e^{0}=1 & x = 0\end{cases}$

 Now,         $(hof)'(0)=\lim_{h \rightarrow0}\frac{e^{|g(x)|}-1}{x}$

                        $=\lim_{h \rightarrow0}\frac{e^{|g(x)|}-1}{|g(x)|}.\frac{|g(x)|}{x}$

                $=\lim_{h \rightarrow0}\frac{e^{|g(x)|}-1}{|g(x)|}.\lim_{h \rightarrow 0}\frac{|g(x)-0|}{|x|}.\lim_{h \rightarrow 0}\frac{|x|}{x}$

                            = $1.g'(0).\lim_{h \rightarrow 0}\frac{|x|}{x}$

    = 0, as g'(0)=0

     $\therefore$      Option  (d) is correct.