# Composition of piecewise functions – Strange result

algebra-precalculusfunction-and-relation-compositionfunctionssolution-verification

I'm trying to get $$f(g(x))$$, where:

$$f(x)= \begin{cases} \sqrt{1-x} &\text{if } x \leq 1 \\ x &\text{if } x > 1 \end{cases}$$

$$g(x)= \begin{cases} x + 1 &\text{if } x \geq 0 \\ x^2 &\text{if } x < 0 \end{cases}$$

I
followed these steps:

1. $$g(x \geq 0) = x + 1$$ hence:
$$f(g(x \geq 0)) = \begin{cases} \sqrt{1 – (x + 1)} = \sqrt{-x} &\text{if } 0 \leq x \leq 1 \\ x + 1 &\text{if } x > 1 \end{cases}$$

but $$\sqrt{-x}$$ is not real! Is it correct? if not, what is the right result?

1. $$g(x < 0) = x^2$$ hence:

$$f(g(x < 0)) = \sqrt{1 – x^2} \quad \text{if } x < 0$$

EDIT: steps to get the @JoséCarlosSantos solution

$$\boldsymbol{g(x) \leq 1}$$

$$\begin{cases} x + 1 \leq 1 \\ x^2 \leq 1 \end{cases}$$

$$\begin{cases} x \leq 0 \\ -1 \leq x \leq 1 \end{cases}$$

so $$-1 \leq x \leq 0$$.

$$\boldsymbol{g(x) > 1}$$

$$\begin{cases} x + 1 > 1 \\ x^2 > 1 \end{cases}$$

$$\begin{cases} x > 0 \\ x < – 1 \text{ or } x > 1 \end{cases}$$

this means that I can split it in two subsystems:

1. $$\begin{cases} x > 0 \\ x < – 1 \end{cases}$$ this system does not have any solution;
2. $$\begin{cases} x > 0 \\ x > 1 \end{cases}$$ the solution is $$x > 1$$.

Putting together the results of the two subsystems: $$x > 1$$. I made a mistake, the right result should be $$x < – 1$$ or $$x > 0$$.

Since$$g(x)=\begin{cases}x+1&\text{ if }x\geqslant0\\x^2&\text{ if }x<0,\end{cases}$$you have$$g(x)\leqslant1\text{ if }x\in[-1,0]\quad\text{and}\quad g(x)>1\text{ if }x<-1\text{ or }x>0.$$So,\begin{align}f\bigl(g(x)\bigr)&=\begin{cases}\sqrt{1-g(x)}&\text{ if }x\in[-1,0]\\g(x)&\text{ if }x<-1\text{ or }x>0\end{cases}\\&=\begin{cases}\sqrt{1-x^2}&\text{ if }x\in[-1,0)\\0&\text{ if }x=0\\x^2&\text{ if }x<-1\\x+1&\text{ if }x>0.\end{cases}\end{align}