Question

1. - / 4 points evaluate f(-4), f(0), and f(3) for the piecewise defined function.

$f(x) = \

$$\begin{cases} x^2 - 5 & \\text{if } x < 0 \\\\ x & \\text{if } x \\geq 0 \\end{cases}$$

$
$f(-4) = \square$
$f(0) = \square$
$f(3) = \square$
sketch the graph of the function.

Explanation:

Step1: Evaluate $f(-4)$ (use $x<0$ rule)

Since $-4 < 0$, use $f(x)=x^2-5$.

$$\begin{align*} f(-4) &= (-4)^2 - 5 \\ &= 16 - 5 \\ &= 11 \end{align*}$$

Step2: Evaluate $f(0)$ (use $x\geq0$ rule)

Since $0 \geq 0$, use $f(x)=x$.
$f(0) = 0$

Step3: Evaluate $f(3)$ (use $x\geq0$ rule)

Since $3 \geq 0$, use $f(x)=x$.
$f(3) = 3$

Answer:

$f(-4) = 11$
$f(0) = 0$
$f(3) = 3$

Graph Notes (for sketching):

• For $x<0$: Plot the parabola $y=x^2-5$, with an open circle at $(0, -5)$ (since $x=0$ is not included in this piece).
• For $x\geq0$: Plot the line $y=x$, with a closed circle at $(0, 0)$ (since $x=0$ is included here), extending infinitely upward to the right.