Let f and g be two functions defined by f(x)=casesx+1, & x

Question

Let $f$ and $g$ be two functions defined by $f(x)=\begin{cases}x+1, & x<0\ |x-1|, & x\ge0\end{cases}$ and $g(x)=\begin{cases}x+1, & x<0\ 1, & x\ge0\end{cases}$. Then $(g\circ f)(x)$ is:

Accepted Answer

Rewriting $f$: $f(x)=\begin{cases}x+1, & x<0\ 1-x, & 0\le x<1\ x-1, & x\ge1\end{cases}$

$$g(f(x))=\begin{cases}f(x)+1, & f(x)<0\ 1, & f(x)\ge0\end{cases}$$

$f(x)<0$ only when $x+1<0$, i.e. $x<-1$ (checking each branch, this is the only region where $f(x)$ is negative). So
$$g(f(x))=\begin{cases}x+2, & x<-1\ 1, & x\ge-1\end{cases}$$

This is continuous everywhere (both pieces agree at $x=-1$: $x+2\to1$), but the derivative jumps from $1$ (left) to $0$ (right) at $x=-1$, so $g\circ f$ is not differentiable exactly at $x=-1$ --- one point.

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