Question

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give the intervals of increase and decrease for the graph above.
warning: there may not be two intervals for each part. if you dont need an answer box leave blank.
(a) increasing on the interval(s)
and
(b) decreasing on the interval(s)
and
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Explanation:

Step1: Analyze the graph's increasing intervals

A function is increasing when as \( x \) increases, \( y \) also increases. Looking at the graph, we observe the following:
- From \( x = -2 \) to \( x = 1 \)? Wait, no, let's re - examine. Wait, the left - most part: from \( x=-3 \) (the peak? Wait, no, the graph has a peak at \( x = - 2 \), then a valley at \( x = 1 \), and then increases again. Wait, actually, let's find the critical points. The graph has a local maximum at \( x=-2 \), a local minimum at \( x = 1 \). So, for increasing:
- To the left of \( x=-2 \), as \( x \) increases towards \( x=-2 \), the function increases? Wait, no. Wait, when \( x \) goes from \( - \infty \) (but in the graph, the domain shown is from - 5 to 5). Wait, looking at the graph, from \( x=-2 \) to \( x = 1 \), the function is decreasing? No, wait, let's look at the slope. When \( x \) increases from \( - 3 \) to \( - 2 \), the function goes up (increasing). Then from \( x=-2 \) to \( x = 1 \), the function goes down (decreasing). Then from \( x = 1 \) to \( x=\infty \) (in the graph, up to \( x = 5 \)), the function goes up (increasing). Wait, but in the graph, the left - hand side: the graph comes from below, rises to a peak at \( x=-2 \), then falls to a valley at \( x = 1 \), then rises again. So the increasing intervals are \( (-2,1) \)? No, wait, no. Wait, when \( x \) increases, if the function's \( y \) - value increases, it's increasing. So:
- From \( x=-3 \) to \( x=-2 \): as \( x \) increases from - 3 to - 2, \( y \) increases (so interval \( (-3,-2) \)? No, wait, the peak is at \( x=-2 \). Wait, maybe I made a mistake. Let's look at the x - axis. The graph has a root at \( x=-3 \), then rises to a maximum at \( x=-2 \), then falls, crosses the x - axis at \( x = 0 \), then falls to a minimum at \( x = 1 \), then rises, crossing the x - axis at \( x = 2 \), then rising. So:
- Increasing intervals: when the function's slope is positive. So from \( x=-2 \) to \( x = 1 \)? No, slope is negative there. Wait, no:
- From \( x=-\infty \) (in the graph, from \( x=-5 \) to \( x=-2 \)): as \( x \) increases from - 5 to - 2, the function is increasing (since it goes from a low point to the peak at \( x=-2 \)). Then from \( x = 1 \) to \( x = 5 \) (in the graph), the function is increasing (from the valley at \( x = 1 \) to the right - hand side). Wait, no, the valley is at \( x = 1 \), so after \( x = 1 \), as \( x \) increases, \( y \) increases. And before \( x=-2 \), as \( x \) increases towards \( x=-2 \), \( y \) increases. So the increasing intervals are \( (-2,1) \)? No, that's decreasing. Wait, I'm confused. Let's use the definition: a function \( f(x) \) is increasing on an interval \( (a,b) \) if for any \( x_1,x_2\in(a,b) \) with \( x_1<x_2 \), \( f(x_1)<f(x_2) \).
- Looking at the graph:
- From \( x=-2 \) to \( x = 1 \): if \( x_1=-2 \) and \( x_2 = 1 \), \( f(-2)>f(1) \), so it's decreasing.
- From \( x = 1 \) to \( x = 5 \): as \( x \) increases from 1 to 5, \( f(x) \) increases (so interval \( (1,+\infty) \), but in the graph, up to \( x = 5 \), so \( (1,5) \) or \( (1,\infty) \) in the domain shown).
- From \( x=-5 \) to \( x=-2 \): as \( x \) increases from - 5 to - 2, \( f(x) \) increases (since it goes from a lower value to the peak at \( x=-2 \)), so interval \( (-5,-2) \)? Wait, but the graph starts from below at \( x=-5 \), rises to \( x=-2 \). So the increasing intervals are \( (-2,1) \)? No, that's decreasing. Wait, I think I messed up the critical points. Let's re - express:
- The local maximum is at \( x=-2 \), local minimum at \( x = 1 \). So the function is increasing when \( x < - 2 \) (moving from left to right towards \( x=-2 \)) and when \( x>1 \) (moving from left to right away from \( x = 1 \)). So the increasing intervals are \( (-\infty,-2) \) (but in the graph, the domain shown is from - 5 to 5, so \( (-5,-2) \) is not correct. Wait, the graph's left - most part: at \( x=-4 \), the function is low, then rises to \( x=-2 \) (peak), then falls to \( x = 1 \) (valley), then rises. So the increasing intervals are \( (-2,1) \)? No, that's decreasing. Wait, I think I have the direction wrong. Let's take two points:
- For \( x=-3 \) and \( x=-2 \): \( f(-3)<f(-2) \), so increasing on \( (-3,-2) \)? No, \( x=-3 \) to \( x=-2 \): \( x \) increases, \( f(x) \) increases. Then \( x=-2 \) to \( x = 1 \): \( x \) increases, \( f(x) \) decreases (since \( f(-2)>f(1) \)). Then \( x = 1 \) to \( x = 3 \): \( x \) increases, \( f(x) \) increases (since \( f(1)<f(3) \)). So the increasing intervals are \( (-3,-2) \) and \( (1,3) \)? Wait, no, the graph at \( x = 2 \) is crossing the x - axis, and then going up. So the correct increasing intervals are \( (-2,1) \)? No, I'm really confused. Wait, let's look at the standard way: the function is increasing when the graph is going up as we move from left to right. So:
- From \( x=-2 \) to \( x = 1 \): the graph is going down (decreasing).
- From \( x = 1 \) to \( x = 5 \): the graph is going up (increasing).
- From \( x=-5 \) to \( x=-2 \): the graph is going up (increasing). So the increasing intervals are \( (-\infty,-2) \) and \( (1,\infty) \) (in the context of the graph, the visible part: \( (-5,-2) \) and \( (1,5) \), but since the problem doesn't specify the domain beyond what's shown, but in general, the critical points are at \( x=-2 \) (local max) and \( x = 1 \) (local min). So the increasing intervals are \( (-2,1) \)? No, that's decreasing. Wait, I think I had the local max and min wrong. The local maximum is at \( x=-2 \), local minimum at \( x = 1 \). So when \( x < - 2 \), as \( x \) increases, \( f(x) \) increases (towards the local max). When \( x>1 \), as \( x \) increases, \( f(x) \) increases (away from the local min). When \( -2<x<1 \), as \( x \) increases, \( f(x) \) decreases (towards the local min). So the increasing intervals are \( (-\infty,-2) \) and \( (1,\infty) \). In the graph's visible range (from - 5 to 5), the increasing intervals are \( (-5,-2) \) and \( (1,5) \). But maybe the problem expects the intervals based on the critical points \( x=-2 \) and \( x = 1 \). So the increasing intervals are \( (-\infty,-2) \) and \( (1,\infty) \).

Step2: Analyze the graph's decreasing intervals

A function is decreasing when as \( x \) increases, \( y \) decreases. Using the critical points \( x=-2 \) (local max) and \( x = 1 \) (local min), the function is decreasing on the interval between the local max and local min, i.e., \( (-2,1) \).

Answer:

(a) Increasing on the interval(s) \((-\infty, - 2)\) and \((1,\infty)\) (or in the graph's visible domain, \((-5,-2)\) and \((1,5)\))
(b) Decreasing on the interval(s) \((-2,1)\)