## Precalculus: Mathematics for Calculus, 7th Edition

$f(-2)=4$ $f(-1)=1$ $f(0)=1$ $f(1)=2$ $f(2)=3$
$f(x)=\begin{cases}x^{2} \mbox {if }x\lt0\\x+1 \mbox {if }x\ge0\end{cases}$ $f(-2),$ $f(-1),$ $f(0),$ $f(1),$ $f(2)$ $f(-2)$ Since $-2\lt0$, substitute $x$ by $-2$ in $x^{2}$ to find $f(-2)$: $f(-2)=(-2)^{2}=4$ $f(-1)$ Since $-1\lt0$, substitute $x$ by $-1$ in $x^{2}$ to find $f(-1)$: $f(-1)=(-1)^{2}=1$ $f(0)$ Since $0\ge0$, substitute $x$ by $0$ in $x+1$ to find $f(0)$: $f(0)=0+1=1$ $f(1)$ Since $1\ge0$, substitute $x$ by $1$ in $x+1$ to find $f(1)$: $f(1)=1+1=2$ $f(2)$ Since $2\ge0$, substitute $x$ by $2$ in $x+1$ to find $f(2)$: $f(2)=2+1=3$