Mathematics - Continuity - Differentiability Question with Solution | TestHub

Here $f\left(x\right)=\left\{\begin{array}{lll}x+3& ;& x<-3\\ -\left(x+3\right)& ;& -3\le x<0\\ e^x& ;& x\ge 0\end{array}\right\}$

and $g\left(x\right)=\left\{\begin{array}{lll}{x}^2+k_{1}x& ;& x<0\\ 4x+k_2& ;& x\ge 0\end{array}\right\}$

Now $g\left(f\left(x\right)\right)=\left\{\begin{array}{lll}f{\left(x\right)}^2+k_{1}f\left(x\right)& ;& f\left(x\right)<0\\ 4f\left(x\right)+k_2& ;& f\left(x\right)\ge 0\end{array}\right\}$

i.e. $g\left(f\left(x\right)\right)=\left\{\begin{array}{ccc}{\left(x+3\right)}^2+k_1\left(x+3\right)& ;& x<-3\\ {\left(x+3\right)}^2-k_1\left(x+3\right)& ;& -3\le x<0\\ 4e^x+k_2& ;& x\ge 0\end{array}\right\}$

For continuity at $x=0$, 

$\mathit{gof}\left(0\right)=g\left(f\left(0^{-}\right)\right)=g\left(f\left(0^{+}\right)\right)$

i.e. $4+k_2=9-3k_1=4+k_2$

$\Rightarrow 3k_1+k_2=5 \cdots \left(i\right)$

Now differentiating, we get

${\left(g\left(f\left(x\right)\right)\right)}^{\text{'}}=\left\{\begin{array}{ccc}2\left(x+3\right)+k_1& ;& x<-3\\ 2\left(x+3\right)-k_1& ;& -3\le x<0\\ 4e^x& ;& x\ge 0\end{array}\right\}$

At $x=0$,

$6-k_1=4$

$\Rightarrow k_1=2 \dots \left(ii\right)$

$\therefore k_1=2, k_2=-1$ (from $\left(i\right)$)

So $\mathit{gof}\left(x\right)=\left\{\begin{array}{ccc}{\left(x+3\right)}^2+2\left(x+3\right)& ;& x<-3\\ {\left(x+3\right)}^2-2\left(x+3\right)& ;& -3\le x<0\\ 4e^x-1& ;& x\ge 0\end{array}\right\}$

Hence, $\mathit{gof}\left(-4\right)+\mathit{gof}\left(4\right)=4e^4-2$$=2\left(2e^4-1\right)$