Functions — Question 6

Question 6 - Solution

(a) The absolute values change sign at $x=-1$ and $x=2$.

$$f(x)= \begin{cases} -(x-2)-(x+1), & x<-1, \\[0.3em] -(x-2)+(x+1), & -1\leq x<2, \\[0.3em] (x-2)+(x+1), & x\geq 2. \end{cases}$$

After simplifying,

$$f(x)= \begin{cases} -2x+1, & x<-1, \\[0.4em] 3, & -1\leq x<2, \\[0.4em] 2x-1, & x\geq 2. \end{cases}$$

(b) The graph decreases until $x=-1$, stays constant from $x=-1$ to $x=2$, and then increases.

(c) The only possible non-differentiable points are where the absolute value expressions change sign:

$$x=-1 \qquad \text{and} \qquad x=2.$$

From the piecewise function, the slopes are

$$\begin{cases} -2, & x<-1, \\ 0, & -1<x<2, \\ 2, & x>2. \end{cases}$$

At $x=-1$, the slope changes from $-2$ to $0$, so $f$ is not differentiable there.

At $x=2$, the slope changes from $0$ to $2$, so $f$ is not differentiable there.

Therefore,

$$\boxed{x=-1,\ 2}$$