f, g: R → R be two real valued function defined as f(x) = \ cases -...

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

and g\left(x\right) = \left\{\begin{cases}{x}^{2} + {k}_{1}x & ; & x < 0 \\ 4x + {k}_{2} & ; & x \ge 0\end{cases}\right\}

Now g\left(f\left(x\right)\right) = \left\{\begin{cases}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{cases}\right\}

i.e. g\left(f\left(x\right)\right) = \left\{\begin{cases}{\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 \\ 4{e}^{x} + {k}_{2} & ; & x \ge 0\end{cases}\right\}

For continuity at $x = 0$,

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

i.e. $4 + {k}_{2} = 9 - 3{k}_{1} = 4 + {k}_{2}$

$⇒ 3{k}_{1} + {k}_{2} = 5 ⋯ \left(i\right)$

Now differentiating, we get

{\left(g\left(f\left(x\right)\right)\right)}^{'} = \left\{\begin{cases}2\left(x + 3\right) + {k}_{1} & ; & x < - 3 \\ 2\left(x + 3\right) - {k}_{1} & ; & - 3 \le x < 0 \\ 4{e}^{x} & ; & x \ge 0\end{cases}\right\}

At $x = 0$,

$6 - {k}_{1} = 4$

$⇒ {k}_{1} = 2 \ldots \left(ii\right)$

$∴ {k}_{1} = 2, {k}_{2} = - 1$ (from \left(i\right)$)</p><p>So$gof \left(x\right) = \left\{$\begin{cases}{\left(x + 3\right)}^{2} + 2\left(x + 3\right) & ; & x < - 3 \\ {\left(x + 3\right)}^{2} - 2\left(x + 3\right) & ; & - 3 \le x < 0 \\ 4{e}^{x} - 1 & ; & x \ge 0\end{cases}$\right\}$</p><p>Hence,$gof \left(- 4\right) + gof \left(4\right) = 4{e}^{4} - 2$$= 2\left(2{e}^{4} - 1\right)