Question

the graph of a function g is given.
(a) state the values of the following.
$g(3) = $
$g(1) = $
$g(4) = 0$
$g(2) = $
(b) for what value(s) of x is $g(x) = 3$? (enter your answers as a comma - separated list.)
(c) for what value(s) of x is $g(x) \leq 3$? (enter your answer using interval notation.)

Explanation:

Step1: Find g(3) from the graph

Locate $x=3$, the filled point has $y=1$? No, correction: At $x=3$, the solid point is at $y=1$? No, recheck: The graph at $x=3$ has a solid red dot at $y=1$? Wait no, no—wait $x=3$: the right curve has a solid dot at $x=3$, $y=1$. Wait no, $x=2$: open dot at $y=3$, solid dot at $x=2$? No, $x=1$: solid dot at $y=1$. Wait, let's re-express each:

Step1: Evaluate g(3)

At $x=3$, the solid point gives $g(3)=1$? No, wait no—wait the graph: left curve starts at $x=-4, y=3$ (solid), goes down to $x=0, y=-2$ (solid), then right curve goes up to $x=2, y=1$ (solid), then up to $x=2, y=3$ (open), then down to $x=4, y=0$ (solid).
Wait, correction:

Step1: Find g(3)

Locate $x=3$ on the graph: the curve has $y$ value between 1 and 3? No, no—wait $x=3$: the right segment from open $(2,3)$ to solid $(4,0)$: at $x=3$, $g(3)$ is the y-value on that line. The line from $(2,3)$ (open) to $(4,0)$: slope is $\frac{0-3}{4-2}=-\frac{3}{2}$. Equation: $y-0=-\frac{3}{2}(x-4)$, so at $x=3$: $y=-\frac{3}{2}(3-4)=\frac{3}{2}$? No, wait no, the left curve: from $(-4,3)$ (solid) to $(0,-2)$ (solid): that's a parabola? Wait no, the problem's graph:
Wait, let's do each part correctly:

Part (a)

Step1: Calculate g(3)

At $x=3$, the right curve (from open $(2,3)$ to solid $(4,0)$) has $g(3)=\frac{3}{2}$? No, no—wait no, the solid dot at $x=2$ is $y=1$, open at $x=2$ is $y=3$. So $x=3$: the point on the curve between $(2,3)$ (open) and $(4,0)$ (solid): when $x=3$, $y=\frac{3}{2}$? No, wait no, maybe it's a parabola. Wait no, the user's wrong answers: $g(3)=1$ was wrong, $g(1)=1$ wrong, $g(2)=3$ wrong.
Wait, $g(1)$: at $x=1$, the right curve (from $(0,-2)$ to $(2,1)$) is a parabola? At $x=1$, $y$ is $\frac{(-2 + 1)}{2}$? No, $(0,-2)$ to $(2,1)$: slope $\frac{3}{2}$, equation $y=\frac{3}{2}x -2$. At $x=1$, $y=\frac{3}{2}(1)-2=-\frac{1}{2}$? No, that can't be. Wait no, the graph's lowest point is $(0,-2)$ (solid). So left curve: $(-4,3)$ to $(0,-2)$ (parabola), right curve: $(0,-2)$ to $(2,1)$ (parabola), then $(2,3)$ (open) to $(4,0)$ (line).
Ah, right! The right side has two parts: from $(0,-2)$ to $(2,1)$ (solid, so $x$ from 0 to 2, solid at $(2,1)$), then from $(2,3)$ (open, so $x>2$) to $(4,0)$ (solid).
So:

Step1: Find g(3)

$x=3$ is on $(2,3)$ to $(4,0)$: line equation $y = -\frac{3}{2}x +6$. At $x=3$: $y=-\frac{9}{2}+6=\frac{3}{2}$? No, wait when $x=2$, $y=3$ (open): $-\frac{3}{2}(2)+6=3$, correct. $x=4$: $-\frac{3}{2}(4)+6=0$, correct. So $g(3)=\frac{3}{2}$? No, but the user's answer 1 was wrong. Wait no, maybe the graph is symmetric? Left curve $(-4,3)$ to $(0,-2)$, right curve $(0,-2)$ to $(4,3)$ but with open at $(2,3)$? No, the solid dot at $(4,0)$.
Wait, no, let's look at the user's correct answer: $g(4)=0$ is correct. So $x=4$, $y=0$.
$g(2)$: the solid dot is at $y=1$, open at $y=3$, so $g(2)=1$.
$g(1)$: on the curve from $(0,-2)$ to $(2,1)$: at $x=1$, $y=\frac{-2+1}{2}$? No, it's a parabola with vertex at $(0,-2)$, so equation $y=ax^2-2$. At $x=-4$, $y=3$: $3=16a-2 \to 16a=5 \to a=\frac{5}{16}$. So $g(1)=\frac{5}{16}(1)^2-2=-\frac{27}{16}$? No, that can't be. Wait no, the left curve from $(-4,3)$ to $(0,-2)$ is a straight line? Slope $\frac{-2-3}{0-(-4)}=-\frac{5}{4}$. Equation: $y=-\frac{5}{4}x -2$. At $x=-4$, $y=-\frac{5}{4}(-4)-2=5-2=3$, correct. At $x=0$, $y=-2$, correct.
Right curve from $(0,-2)$ to $(2,1)$: slope $\frac{1-(-2)}{2-0}=\frac{3}{2}$. Equation: $y=\frac{3}{2}x -2$. At $x=2$, $y=\frac{3}{2}(2)-2=1$, correct. Then from $(2,3)$ to $(4,0)$: slope $\frac{0-3}{4-2}=-\frac{3}{2}$. Equation: $y=-\frac{3}{2}(x-4)=-\frac{3}{2}x +6$. At $x=2$, $y=3$, correct.
Now:

Step1: Compute g(3)

Use line equation for $x>2$: $g(3)=-\frac{3}{2}(3)+6 = \frac{3}{2}$

Step2: Compute g(1)

Use line equation for $0\leq x\leq2$: $g(1)=\frac{3}{2}(1)-2 = -\frac{1}{2}$

Step3: Compute g(2)

Use solid point at $x=2$: $g(2)=1$

Step4: Confirm g(4)

Solid point at $x=4$: $g(4)=0$ (already correct)

Part (b)

Step5: Find x where g(x)=3

Look for $y=3$: solid point at $x=-4$, open point at $x=2$ (so $x=2$ is not included). So $x=-4$
Wait no, left curve: line from $(-4,3)$ to $(0,-2)$: $y=3$ only at $x=-4$. Right curve: open point at $(2,3)$, so $x=2$ is not in domain for that value. So $x=-4$

Part (c)

Already correct: $[-4,4]$

Step1: Find g(3)

Use line for $x>2$: $g(3)=-\frac{3}{2}(3)+6=\frac{3}{2}$

Step2: Find g(1)

Use line for $0\leq x\leq2$: $g(1)=\frac{3}{2}(1)-2=-\frac{1}{2}$

Step3: Find g(2)

Use solid point at $x=2$: $g(2)=1$

Step4: Confirm g(4)

Solid point at $x=4$: $g(4)=0$

Step5: Solve g(x)=3

Locate $y=3$: only solid point at $x=-4$

Step6: Confirm g(x)≤3

Interval is all valid $x$: $[-4,4]$

Answer:

(a)
$g(3)=\frac{3}{2}$
$g(1)=-\frac{1}{2}$
$g(4)=0$
$g(2)=1$

(b)
$-4$

(c)
$[-4,4]$

Wait, no, wait the left curve: is $x=-4$ the only point with $g(x)=3$? Yes, because the left line only hits $y=3$ at $x=-4$, and the right side has an open point at $x=2, y=3$, so $x=2$ does not satisfy $g(x)=3$.

Wait, maybe I misread the graph: maybe the right curve's open point is at $x=2, y=3$, and the left curve is a parabola that goes from $(-4,3)$ to $(0,-2)$ to $(4,3)$, but with a solid dot at $(4,0)$? No, the user's $g(4)=0$ is correct, so $x=4$ has $y=0$, not 3.

Correcting part (b): the only $x$ where $g(x)=3$ is $x=-4$, because the open circle at $x=2$ means $g(2)
eq3$, $g(2)=1$.

Now reformatting properly: