Key Concepts and Formulas

• Continuity of a function: A function $h(x)$ is continuous at a point $c$ if $\lim_{x \to c^-} h(x) = \lim_{x \to c^+} h(x) = h(c)$. For piecewise functions, continuity at the point where the definition changes requires the left-hand limit, right-hand limit, and the function value at that point to be equal.
• Composition of functions: For two functions $f$ and $g$, the composite function $(g \circ f)(x)$ is defined as $g(f(x))$. Similarly, $(f \circ g)(x)$ is defined as $f(g(x))$.
• Absolute value function: $|y|$ is equal to $y$ if $y \ge 0$, and $-y$ if $y < 0$.

Step-by-Step Solution

Step 1: Determine the values of $a$ and $b$ using the continuity of $f(x)$ and $g(x)$.

The problem states that both $f(x)$ and $g(x)$ are continuous on $\mathbb{R}$.

For $f(x)$ to be continuous at $x=0$: The left-hand limit is $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x+a) = 0+a = a$. The right-hand limit is $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} |x-4|$. Since $x > 0$, as $x \to 0^+$, $x-4$ approaches $-4$. Thus, $|x-4|$ approaches $|-4| = 4$. The function value at $x=0$ is $f(0) = 0+a = a$. For continuity at $x=0$, we must have $a = 4$.

For $g(x)$ to be continuous at $x=0$: The left-hand limit is $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (x+1) = 0+1 = 1$. The right-hand limit is $\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} ((x-4)^2 + b) = (0-4)^2 + b = 16+b$. The function value at $x=0$ is $g(0) = (0-4)^2 + b = 16+b$. For continuity at $x=0$, we must have $1 = 16+b$, which implies $b = 1 - 16 = -15$.

So, we have $a=4$ and $b=-15$.

Step 2: Rewrite the functions $f(x)$ and $g(x)$ with the determined values of $a$ and $b$.

With $a=4$ and $b=-15$, the functions become: f(x) = \left\{ {\matrix{ {x + 4} & , & {x \le 0} \cr {|x - 4|} & , & {x > 0} \cr } } \right. g(x) = \left\{ {\matrix{ {x + 1} & , & {x < 0} \cr {{{(x - 4)}^2} - 15} & , & {x \ge 0} \cr } } \right.

Step 3: Calculate $(g \circ f)(2)$.

First, we need to find $f(2)$. Since $2 > 0$, we use the second case for $f(x)$: $f(2) = |2 - 4| = |-2| = 2$.

Now, we need to find $g(f(2))$, which is $g(2)$. Since $2 \ge 0$, we use the second case for $g(x)$: $g(2) = (2 - 4)^2 - 15 = (-2)^2 - 15 = 4 - 15 = -11$. Therefore, $(g \circ f)(2) = -11$.

Step 4: Calculate $(f \circ g)(-2)$.

First, we need to find $g(-2)$. Since $-2 < 0$, we use the first case for $g(x)$: $g(-2) = -2 + 1 = -1$.

Now, we need to find $f(g(-2))$, which is $f(-1)$. Since $-1 \le 0$, we use the first case for $f(x)$: $f(-1) = -1 + 4 = 3$. Therefore, $(f \circ g)(-2) = 3$.

Step 5: Calculate $(g \circ f)(2) + (f \circ g)(-2)$.

Using the results from Step 3 and Step 4: $(g \circ f)(2) + (f \circ g)(-2) = -11 + 3 = -8$.

Common Mistakes & Tips

• Incorrectly evaluating absolute values: Be careful when evaluating $|x-4|$ for $x>0$. For example, when $x=2$, $|2-4| = |-2| = 2$, not $2-4$.
• Confusing the order of composition: Remember that $(g \circ f)(x) = g(f(x))$, meaning you evaluate the inner function first.
• Checking the correct piecewise definition: Always ensure you are using the correct part of the piecewise function definition based on the input value to the function.

Summary

The problem requires us to first find the unknown constants $a$ and $b$ by using the condition that the given piecewise functions $f(x)$ and $g(x)$ are continuous on $\mathbb{R}$. We applied the definition of continuity at the point where the function definition changes ($x=0$) for both $f(x)$ and $g(x)$ to find $a=4$ and $b=-15$. Then, we calculated the values of the composite functions $(g \circ f)(2)$ and $(f \circ g)(-2)$ by carefully evaluating the inner and outer functions using their respective piecewise definitions. Finally, we summed these two values to get the required result.

The final answer is \boxed{-8}.