Question

refer to the graph of $y = f(x)$ provided to determine the interval(s), if any, where $f < 0$.
(description: curve rises from the 3rd quadrant up through the x-axis near 5 into the 2nd quadrant, turns downward at x =3, continues to fall through the origin in an upward arc shape, turning to rise at x = 3 and rising up into the 1st quadrant, interesting the x axis near 5.)
\bigcirc no correct answer choice is given.
\bigcirc $(-\infty, -3) \cup (3, \infty)$
\bigcirc $(-3, 3)$
\bigcirc $(0, \infty)$
\bigcirc $(-\infty, -5) \cup (0, 5)$

Explanation:

Step1: Recall derivative sign meaning

$f'(x) < 0$ means the function $f(x)$ is decreasing (slope of tangent is negative).

Step2: Analyze graph's increasing/decreasing

From the description: The curve rises (increasing) until $x = -3$ (turns downward), then falls (decreasing) until $x = 3$ (turns to rise), then rises (increasing) after $x = 3$. Wait, no—wait the description: "turns downward at $x = -3$ (wait, original description: "turns downward at x =3"? Wait, re-reading: "Curve rises from the 3rd quadrant up through the x - axis near -5 (maybe typo, near -5?) into the 2nd quadrant, turns downward at x = -3 (corrected?), continues to fall through the origin in an upward arc shape, turning to rise at x = 3 and rising up into the 1st quadrant, intersecting the x axis near 5." Wait, maybe the peak is at $x=-3$, then it decreases until $x = 3$ (valley), then increases. Wait, no—if it rises to $x=-3$, then falls until $x = 3$, then rises. So when is $f'(x)<0$? When the function is decreasing. So from $x=-3$ to $x = 3$, the function is decreasing (since it goes from peak at $x=-3$ down to valley at $x = 3$). Wait, but the options: one option is $(-3,3)$. Wait, let's check the options. Wait, maybe the peak is at $x=-3$, then it decreases until $x = 3$, so the interval where $f'(x)<0$ is $(-3,3)$, because in that interval, the function is decreasing (slope negative). Let's check the options: option C is $(-3,3)$. Let's verify: when $x < -3$, the function is increasing (so $f'(x)>0$), at $x=-3$, it's a local max (so $f'(-3)=0$), then from $x=-3$ to $x = 3$, the function is decreasing (so $f'(x)<0$), at $x = 3$, local min ($f'(3)=0$), then for $x>3$, increasing ($f'(x)>0$). So the interval where $f'(x)<0$ is $(-3,3)$.

Answer:

\( (-3, 3) \) (corresponding to the option "(-3, 3)")