Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f(x)=\{\begin{array}{ll}x+a,& x\le 0\\ |x-4|,& x>0\end{array}$ and $g(x)=\{\begin{array}{ll}x+1,& x<0\\ (x-4{)}^2+b,& x\ge 0\end{array}$ are continuous on $\mathbb{R}$, then $(g\circ f)(2)+(f\circ g)(-2)$ is equal to :

Imagine you are standing at a crossroads. To your left, the road is paved with one set of rules; to your right, a completely different landscape emerges. This is exactly what a piecewise function is—a mathematical journey where the rules of the game change depending on where you stand.

In this problem, we are dealing with two such functions, $f(x)$ and $g(x)$, and they are both continuous on the entire set of real numbers $\mathbb{R}$. This continuity is our golden ticket. It tells us that no matter how the rules change at $x=0$, the two paths must meet seamlessly.

The Continuity Bridge

For a function to be continuous at a point, the left-hand limit must equal the right-hand limit. Think of it as a bridge that must be perfectly aligned.

For $f(x)$, the transition point is $x=0$. The left-hand limit is ${\lim }_{x\to 0^{-}}(x+a)=a$. The right-hand limit is ${\lim }_{x\to 0^{+}}|x-4|=|0-4|=4$. Since the function is continuous, these must be equal: $a=4$.

Now, let's turn our attention to $g(x)$. Again, the transition point is $x=0$. The left-hand limit is ${\lim }_{x\to 0^{-}}(x+1)=1$. The right-hand limit is ${\lim }_{x\to 0^{+}}((x-4{)}^2+b)=(0-4{)}^2+b=16+b$.

Setting these equal, we get $1=16+b$, which leads us directly to $b=-15$. We have now fully defined our functions:

The Art of Composition

Now that we have the complete definitions, we need to evaluate $(g\circ f)(2)+(f\circ g)(-2)$. The secret to composite functions is to work from the inside out.

Let's tackle $(g\circ f)(2)$ first. This means we need to find $g(f(2))$.

First, we evaluate $f(2)$. Since $2>0$, we use the second branch of $f(x)$, which is $|x-4|$. So, $f(2)=|2-4|=|-2|=2$.

Now, we take this result and plug it into $g(x)$. We need to find $g(2)$. Since $2\ge 0$, we use the second branch of $g(x)$, which is $(x-4{)}^2-15$.

The first part of our expression is $-11$.

The Final Calculation

Next, we evaluate $(f\circ g)(-2)$, which means $f(g(-2))$. First, we find $g(-2)$. Since $-2<0$, we use the first branch of $g(x)$, which is $x+1$.

Now, we take this result and plug it into $f(x)$. We need to find $f(-1)$. Since $-1\le 0$, we use the first branch of $f(x)$, which is $x+4$.

The second part of our expression is $3$. Finally, we add our two results together:

We have successfully navigated the piecewise landscape and arrived at the final answer. The result is $-8$.