If f(x) = begin{cases} x+a, & x leq 0 |x-4|, | vedclass

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(D) For $f(x)$ to be continuous at $x=0$,$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.$0+a = |0-4| \implies a = 4$.For $g(x)$ to be continuo

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$-8$

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(D) For $f(x)$ to be continuous at $x=0$,$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.$0+a = |0-4| \implies a = 4$.For $g(x)$ to be continuous at $x=0$,$\lim_{x 	o 0^-} g(x) = \lim_{x 	o 0^+} g(x) = g(0)$.$0+1 = (0-4)^2 + b \implies 1 = 16 + b \implies b = -15$.Now,$f(x) = \begin{cases} x+4, & x \leq 0 \ |x-4|, & x > 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x Calculate $(g \circ f)(2) = g(f(2))$. Since $2 > 0$,$f(2) = |2-4| = 2$.$g(2) = (2-4)^2 - 15 = 4 - 15 = -11$.Calculate $(f \circ g)(-2) = f(g(-2))$. Since $-2 $f(-1) = -1+4 = 3$.Therefore,$(g \circ f)(2) + (f \circ g)(-2) = -11 + 3 = -8$.

If $f(x) = \begin{cases} x+a, & x \leq 0 \\ |x-4|, & x > 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x < 0 \\ (x-4)^2+b, & x \geq 0 \end{cases}$ are continuous on $\mathbb{R}$,then $(g \circ f)(2) + (f \circ g)(-2)$ is equal to.

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