Question

based on the graph of ( f(x) ) in the figure, determine where ( f(x) < 0 ). select the correct answer below: o ( (-infty, -5) cup (3, infty) ) o ( (-infty, infty) ) o ( (-infty, -5) cup (-5, 3) ) o ( (-5, 3) )

Explanation:

Step1: Recall derivative sign meaning

$f'(x) < 0$ means the function $f(x)$ is decreasing. So we need to find where the graph of $f(x)$ is decreasing.

Step2: Analyze the graph's slope

Looking at the graph, the function decreases when $x > 3$ (right of the vertex) and increases when $x < -5$? Wait, no, wait. Wait, the graph: let's see the critical points. The vertex (or the point where the slope changes from increasing to decreasing or vice versa). Wait, the graph has a left part and a right part. Wait, actually, the function's derivative being negative (decreasing) occurs when the function is sloping downward. Looking at the graph, the function is decreasing on $(3, \infty)$? No, wait, maybe I misread. Wait, the graph: let's check the x - coordinates. Wait, the options include $(-\infty, -5) \cup (3, \infty)$? Wait, no, wait. Wait, the graph: when $x > 3$, the function is decreasing? Wait, no, maybe the vertex is at $x = 3$? Wait, no, looking at the graph, the function has a minimum or maximum? Wait, the graph is a curve that, on the left side (x < -5) is increasing, between -5 and 3 is decreasing? No, wait, no. Wait, the derivative $f'(x) < 0$ is where the function is decreasing. So we look for the interval where the function's graph is falling (slope negative). From the graph, the function decreases on $(3, \infty)$? No, wait, the options: one of the options is $(-\infty, -5) \cup (3, \infty)$? Wait, no, wait the options are:

1. $(-\infty, -5) \cup (3, \infty)$

2. $(-\infty, \infty)$

3. $(-\infty, -5) \cup (-5, 3)$

4. $(-5, 3)$

Wait, no, the original options:

First option: $(-\infty, -5) \cup (3, \infty)$

Second: $(-\infty, \infty)$

Third: $(-\infty, -5) \cup (-5, 3)$

Fourth: $(-5, 3)$

Wait, actually, when $f'(x) < 0$, the function is decreasing. So we need to find where the graph of $f(x)$ is decreasing. Let's analyze the graph:

The graph of $f(x)$: let's see the critical points (where the slope changes from positive to negative or vice versa). The graph has a "peak" or "valley"? Wait, the graph looks like a cubic? No, it looks like a curve that has a minimum at $x=-5$ and a maximum at $x = 3$? Wait, no, when $x$ increases from $-\infty$ to $-5$, the function is increasing (slope positive), then from $-5$ to $3$, the function is decreasing (slope negative), then from $3$ to $\infty$, the function is increasing (slope positive)? Wait, no, that can't be. Wait, maybe I got the axes wrong. Wait, the x - axis is vertical? No, wait, the standard coordinate system: x - axis horizontal, y - axis vertical. Wait, in the graph, the x - axis is labeled with vertical arrows? Wait, no, maybe the graph is rotated? Wait, no, the user's graph: the x - axis is vertical (labeled with x from -10 to 10, arrows up and down), and y - axis is horizontal (labeled with y from -10 to 10, arrows left and right). Oh! That's the key. So it's a rotated coordinate system: x is vertical, y is horizontal. So the function is $x = f(y)$? Wait, no, the problem says "graph of $f(x)$", so maybe it's a typo, or the axes are swapped. Wait, the problem says "determine where $f'(x) < 0$", so $f(x)$ is a function of x, so x is horizontal, y is vertical. But in the graph, the x - axis is vertical (with arrows up and down, labeled x from -10 to 10), and y - axis is horizontal (arrows left and right, labeled y from -10 to 10). So that means the graph is plotted with x on the vertical axis and y on the horizontal axis. So the function is $x = f(y)$, but the problem says $f(x)$, so maybe it's a mistake, and it's $x = f(y)$, but we need to find where $f'(x) < 0$. Wait, no, maybe the axes are swapped. Let's re - interpret: let's take the horizontal axis as x (even though it's labeled y) and vertical axis as y (even though it's labeled x). So the horizontal axis (labeled y) is x, and vertical axis (labeled x) is y. So the graph is $y = f(x)$, with x on the horizontal (labeled y) and y on the vertical (labeled x). So the graph: when x (horizontal) increases, where is the function decreasing (y - value decreasing). So looking at the graph, the function has a minimum at x=-5 and a maximum at x = 3? Wait, no. Wait, when x (horizontal) is less than -5, the function (vertical y) is increasing as x increases? No, wait, let's look at the graph: the left - most part: as x (horizontal) increases from $-\infty$ to -5, the y (vertical) increases. Then from x=-5 to x = 3, y (vertical) decreases as x increases. Then from x = 3 to $\infty$, y (vertical) increases as x increases. Wait, no, that's not right. Wait, the derivative $f'(x)$ is the slope of the tangent to $y = f(x)$ with respect to x (horizontal). So when the function is decreasing (y decreases as x increases), $f'(x) < 0$. So when does y decrease as x increases? From the graph, when x is between -5 and 3? No, wait, no. Wait, if the horizontal axis is x (labeled y) and vertical is y (labeled x), then:

- For x < -5: as x increases (moves right), y (vertical) increases (moves up) → function is increasing, $f'(x)>0$.

- For -5 < x < 3: as x increases (moves right), y (vertical) decreases (moves down) → function is decreasing, $f'(x)<0$.

- For x > 3: as x increases (moves right), y (vertical) increases (moves up) → function is increasing, $f'(x)>0$.

Wait, but that would mean $f'(x) < 0$ on $(-5, 3)$, but that's one of the options (fourth option: $(-5, 3)$). But wait, maybe I misread the axes. Alternatively, if the vertical axis is x and horizontal is y, then the function is $x = f(y)$, and we need $f'(x) < 0$, but that's more complicated. Wait, the problem says "graph of $f(x)$", so x is the independent variable (horizontal), y is dependent (vertical). So the graph has x - axis horizontal (labeled y) and y - axis vertical (labeled x) is a mislabeling. So correcting the labels: horizontal axis is x (from -10 to 10, left - right), vertical axis is y (from -10 to 10, bottom - top). Then the graph:

- The left part: as x increases (moves right) from $-\infty$ to -5, y increases (moves up) → increasing, $f'(x)>0$.

- Between x=-5 and x = 3: as x increases (moves right), y decreases (moves down) → decreasing, $f'(x)<0$.

- After x = 3: as x increases (moves right), y increases (moves up) → increasing, $f'(x)>0$.

Wait, but that would mean $f'(x) < 0$ on $(-5, 3)$, but the first option is $(-\infty, -5) \cup (3, \infty)$, which would be where $f'(x)>0$? No, I'm confused. Wait, maybe the graph is a function where the derivative is negative when the function is decreasing, and the critical points are at x=-5 and x = 3. Wait, the options:

Option 1: $(-\infty, -5) \cup (3, \infty)$

Option 2: $(-\infty, \infty)$

Option 3: $(-\infty, -5) \cup (-5, 3)$

Option 4: $(-5, 3)$

Wait, maybe I had the slope direction wrong. Let's think about the derivative: $f'(x) < 0$ when the function is decreasing (going down as x increases). If the graph has a maximum at x=-5 and a minimum at x = 3? No, that doesn't make sense. Wait, the graph looks like a "sideways" parabola, but since it's a function $y = f(x)$, it must pass the vertical line test. Wait, the graph as drawn: for a given x (horizontal), there is only one y (vertical), so it is a function. So when x increases, the function first increases, then decreases, then increases? No, the graph: starting from the left (x→ -∞), the function comes from below, rises to a peak at x=-5, then falls to a valley at x = 3, then rises again. Wait, that would mean:

- For x < -5: function is increasing (slope positive), $f'(x)>0$.

- For -5 < x < 3: function is decreasing (slope negative), $f'(x)<0$.

- For x > 3: function is increasing (slope positive), $f'(x)>0$.

But then $f'(x) < 0$ on $(-5, 3)$, which is option 4. But wait, the first option is $(-\infty, -5) \cup (3, \infty)$, which would be where $f'(x)>0$. But maybe I got the peak and valley reversed. If the graph has a valley at x=-5 and a peak at x = 3:

- For x < -5: function is decreasing (slope negative), $f'(x)<0$.

- For -5 < x < 3: function is increasing (slope positive), $f'(x)>0$.

- For x > 3: function is decreasing (slope negative), $f'(x)<0$.

Then $f'(x) < 0$ on $(-\infty, -5) \cup (3, \infty)$, which is option 1. Ah! That must be it. I had the peak and valley reversed. So the graph: when x is very small (left of -5), the function is decreasing (going down as x increases), reaches a minimum at x=-5, then increases until x = 3 (reaches a maximum), then decreases again as x increases beyond 3. So:

- $x < -5$: function decreasing, $f'(x) < 0$.

- $-5 < x < 3$: function increasing, $f'(x) > 0$.

- $x > 3$: function decreasing, $f'(x) < 0$.

Therefore, $f'(x) < 0$ on $(-\infty, -5) \cup (3, \infty)$.

Answer:

$(-\infty, -5) \cup (3, \infty)$ (corresponding to the first option)