## Algebra and Trigonometry 10th Edition

a) 1; b) -1; c) $\begin{cases} -1,\text{ if }x\lt 1\\ 1,\text{ if }x\gt 1 \end{cases}$
We are given the function: $f(x)=\dfrac{|x|}{x}$ a) Compute $f(2)$: $f(2)=\dfrac{|2|}{2}=\dfrac{2}{2}=1$ b) Compute $f(-2)$: $f(-2)=\dfrac{|-2|}{-2}=\dfrac{2}{-2}=-1$ c) Compute $f(x-1)$: $f(x-1)=\dfrac{|x-1|}{x-1}=\begin{cases} \dfrac{-(x-1)}{x-1},\text{ if }x\lt 1\\ \dfrac{x-1}{x-1},\text{ if }x\gt 1 \end{cases}$ $=\begin{cases} -1,\text{ if }x\lt 1\\ 1,\text{ if }x\gt 1 \end{cases}$