Question

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) = ) ( f(0) = ) ( f(3) = ) 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 $f(x)=x$.
$f(3) = 3$

Step4: Verify correct graph

For $x<0$: $f(x)=x^2-5$ is a parabola opening up, shifted down 5, with a closed dot at $x=-5$ (since $x<0$, at $x=0$ it has an open dot: $0^2-5=-5$). For $x\geq0$: $f(x)=x$ is a line through the origin with a closed dot at $x=0$. This matches the top graph (first option).

Answer:

$f(-4) = 11$
$f(0) = 0$
$f(3) = 3$
Correct graph: The top (first) provided graph.