Let f, g: R rightarrow R be two real-valued f | vedclass

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(D) Given $f(x) = \begin{cases} x+3 & x < -3 \ -(x+3) & -3 \leq x < 0 \ e^x & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x^2

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$2(2e^{4}-1)$

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(D) Given $f(x) = \begin{cases} x+3 & x < -3 \ -(x+3) & -3 \leq x < 0 \ e^x & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x^2 + k_1 x & x < 0 \ 4x + k_2 & x \geq 0 \end{cases}$.
For $(g \circ f)(x)$,we analyze the composition:
If $x < -3$,$f(x) = x+3 < 0$,so $(g \circ f)(x) = (x+3)^2 + k_1(x+3)$.
If $-3 \leq x < 0$,$f(x) = -(x+3) < 0$,so $(g \circ f)(x) = (-(x+3))^2 + k_1(-(x+3)) = (x+3)^2 - k_1(x+3)$.
If $x \geq 0$,$f(x) = e^x > 0$,so $(g \circ f)(x) = 4e^x + k_2$.
For differentiability at $x=0$,it must be continuous at $x=0$:
Left-hand limit: $\lim_{x 	o 0^-} (g \circ f)(x) = (0+3)^2 - k_1(0+3) = 9 - 3k_1$.
Right-hand limit: $\lim_{x 	o 0^+} (g \circ f)(x) = 4e^0 + k_2 = 4 + k_2$.
So,$9 - 3k_1 = 4 + k_2 \implies 3k_1 + k_2 = 5$.
Derivative at $x=0$:
Left-hand derivative: $\frac{d}{dx} [(x+3)^2 - k_1(x+3)]_{x=0} = 2(0+3) - k_1 = 6 - k_1$.
Right-hand derivative: $\frac{d}{dx} [4e^x + k_2]_{x=0} = 4e^0 = 4$.
Equating them: $6 - k_1 = 4 \implies k_1 = 2$.
Substituting $k_1=2$ into $3k_1 + k_2 = 5$,we get $6 + k_2 = 5 \implies k_2 = -1$.
Now,$(g \circ f)(-4) = (-4+3)^2 + 2(-4+3) = (-1)^2 + 2(-1) = 1 - 2 = -1$.
$(g \circ f)(4) = 4e^4 + k_2 = 4e^4 - 1$.
Sum: $(g \circ f)(-4) + (g \circ f)(4) = -1 + 4e^4 - 1 = 4e^4 - 2 = 2(2e^4 - 1)$.

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