Combination of a piecewise defined function and a square function

Question

Let's say I have two functions, $f(x)=x^2$, and:

$$g(x)= \begin{cases} - 4 & x \leq 0\\ |x- 4| & x > 0 \end{cases}$$

Now I have to write combinations $f\circ g$ and $g\circ f$.

I think $f\circ g$ should be:

$$f\circ g(x)= \begin{cases} 16 & x \leq 0\\ (x- 4)^2 & x > 0 \end{cases}$$

and $g\circ f$ should be $g\circ f(x)= |x^2- 4|$, considering that $x$ will always be positive, but now I don't get how would I graph these composite functions, and also how to check continuity at $x=0$, especially of $g\circ f$ since I just eliminated half part of it.

Answer 1

$f(x) = x^2$

$$ g(x) = \begin{cases} -4 & \text{if \(x \leq 0\)}\\ |x - 4| & \text{if \(x > 0\)} \end{cases} $$

You correctly found that \begin{align*} (f \circ g)(x) & = f(g(x))\\ & = \begin{cases} f(-4) & \text{if \(x \leq 0\)}\\ f(|x - 4| & \text{if \(x > 0\)} \end{cases} \\ & = \begin{cases} 16 & \text{if \(x \leq 0\)}\\ |x - 4|^2 & \text{if \(x \geq 0\)} \end{cases} \\ & = \begin{cases} 16 & \text{if \(x \leq 0\)}\\ (x - 4)^2 & \text{if \(x \geq 0\)} \end{cases} \end{align*}

Since \begin{align*} \lim_{x \to 0^+} (f \circ g)(x) & = 16\\ \lim_{x \to 0^-} (f \circ g)(x) & = 16\\ \end{align*} we obtain $$\lim_{x \to 0} (f \circ g)(x) = 16$$ For a function to be continuous at $x = 0$, the limit as $x$ approaches $0$ must exist and the function must be equal to its limit when $x = 0$. Since $(f \circ g)(0) = 16$, $$\lim_{x \to 0} (f \circ g)(x) = (f \circ g)(0)$$ Thus, the function $f \circ g$ is continuous at $x = 0$.

Since $x^2 > 0$ unless $x = 0$, where $f(0) = 0$, we obtain \begin{align*} (g \circ f)(x) & = g(f(x))\\ & = g(x^2)\\ & = \begin{cases} -4 & \text{if \(x^2 \leq 0\)}\\ |x^2 - 4| & \text{if \(x^2 > 0\)} \end{cases} \\ & = \begin{cases} -4 & \text{if \(x = 0\)}\\ |x^2 - 4| & \text{otherwise} \end{cases} \end{align*}

Observe that \begin{align*} \lim_{x \to 0^+} (g \circ f)(x) = 4\\ \lim_{x \to 0^-} (g \circ f)(x) = 4 \end{align*} Hence, $$\lim_{x \to 0} (g \circ f)(x) = 4$$ Thus, the limit of the function exists at $x = 0$. However, $(g \circ f)(0) = -4$. Since $$\lim_{x \to 0} (g \circ f)(x) \neq (g \circ f)(0)$$ the function $g \circ f$ is not continuous at $x = 0$ because it is not equal to its limit. Since the continuity could be removed by defining $(g \circ f)(0) = 4$, which would require defining $g(x) = 4$ when $x = 0$, the function $g \circ f$ is said to have a removable discontinuity at $x = 0$.