Finding the domain and range of a difficult piecewise composite function

Question

I recently inquired about finding a formula for a composition of two piecewise functions, but I have been thoroughly confused by a slightly different example. In this case, I have a question about finding the domain and range for a piecewise composite function. The functions are defined as follows:

$$f(x) = \begin{cases} 1-x, & \text{if $x \le 0$} \\ x^{2}, & \text{if $x > 0$} \end{cases}$$

$$g(x) = \begin{cases} -x, & \text{if $x < 1$} \\ x+1, & \text{if $x \ge 1$} \end{cases}$$

Specifically, my question regards how I should find the domain and range of the composite function $f(g(x))$.

My attempts at the problem:

From my gathered understanding of this type of question (credit goes to @tilper and @lulu for this lucid methodology), I should proceed with the idea that $$f(g(x)) = \begin{cases} 1-g(x), & \text{if $g(x) \le 0$} \\ (g(x))^{2}, & \text{if $g(x) > 0$} \end{cases}.$$ Thus, we ascertain when $g(x) \le 0$ and when $g(x) > 0$. For the piece $-x$ of the original $g(x)$ function, we determine that $-x \le 0$ $\implies$ $x \ge 0$. In addition, we determine that $-x > 0$ $\implies$ $x < 0$. However, the condition must be met that x <1, so $g(x)$ is never $\le 0$ on this piece. However, it seems that $g(x) > 0$ on this piece when $x < 0$ $< 1$. Thus, we have established that $g(x) > 0$ when $x < 0$.

On the next piece of $g(x)$, which is $x + 1$, we again determine when $g(x)$ is $\le 0$ and/or $> 0$. $x+1 \le 0$ $\implies$ $x \le -1$, but for this piece the condition must be met that $x \ge 1$, so clearly $g(x)$ is never $\le 0$ on this piece. Next, $x + 1 > 0$ $\implies$ $x > -1$, which eventually satisfies the condition of $x \ge 1$. What do I do about the values for $-1 < x <1$ where this condition is not satisfied? I am absolutely stumped as to how to proceed with this problem and, once I determine the composition, how to find the domain and range of that new composition.

Many thanks to those of you who could see through my rambling.

Answer 1

The best is to draw a table in which the expressions of $g(x)$ and theirs signs are given explicitly: $$\begin{array}{r|p{1.5cm}cp{1.5cm}cp{1.5cm}|} x&&0&&1&\\ \hline g(x)=&-x&0&-x&2&x+1\\ \hline \operatorname{sgn}g(x)& +&|& -&|&+\\ \hline f(g(x))&g(x)^2=x^2&1& \begin{matrix}1-g(x)\\=x+1\end{matrix}&4& (x+1)^2\\ \hline \end{array}$$ From this table, we conclude at once the range of $f\circ g$ is $\mathbf R_+^{*}$.

Answer 2

Not sure if it helps. Let $g$ be any function. Define $f\circ g$ by $$f(g(x)) =\begin{cases} 1-g(x) & \text{ if } g(x)\leq 0 \\ (g(x))^{2} & \text{ if } g(x)>0. \end{cases}$$ Now let $g$ be the function satisfying on your post. We will check $x<1$ and $x\geq 1$ seperately. If $x<1$, then $f(g(x))=1-(-x)=1+x$ for all $1>x\geq 0$; $f(g(x))=(-x)^2=x^2$ for all $x<0\,(<1)$. If $x\geq 1$, then $f(g(x))$ doesn't make sense on the first case; $f(g(x))=(x+1)^2$ for all $x\geq 1\,(>-1)$. Altogether, you have $$f(g(x)) =\begin{cases} x^2 & \text{ if } x<0\\ 1+x & \text{ if } 0\leq x<1 \\ (1+x)^2 & \text{ if } 1\leq x. \end{cases}$$