Question

which interval for the graphed function has a local minimum of 0?
-3,-2
-2,0
1,2
2,4

Explanation:

Step1: Recall local minimum definition

A local minimum on an interval is a point where the function value is less than or equal to neighboring values, and we need the value to be 0.

Step2: Analyze each interval

- For \([-3, -2]\): The point \((-2.5, 0)\) is on the interval, but the local minimum here is \((-1.56, -6)\) (value -6, not 0).
- For \([-2, 0]\): The function goes from \((-2.5, 0)\) down to \((-1.56, -6)\) then up to \((0, 0)\). The local minimum here is -6, not 0.
- For \([1, 2]\): The function is decreasing then increasing? Wait, no, looking at the graph, from \(x = 1.2\) (peak) to \(x = 3\) (the point \((3, 0)\)). Wait, the interval \([2, 4]\): At \(x = 3\), the point is \((3, 0)\), and it's a local minimum (the lowest point in \([2, 4]\) with value 0). Wait, no, let's check the intervals again. Wait the interval \([2, 4]\): The graph has a point at \((3, 0)\) which is a local minimum (since around \(x = 3\), the function is lower or equal? Wait, the graph at \(x = 3\) is a minimum (the bottom of the "valley" there) with \(y = 0\). Wait, but also check \([-3, -2]\): The point \((-2.5, 0)\) is on the interval, but the local minimum in \([-3, -2]\) is the lowest point, which is \((-1.56, -6)\) (but that's in \([-2, 0]\)? Wait, maybe I misread the intervals. Wait the interval \([2, 4]\): The graph from \(x = 2\) to \(x = 4\) has a minimum at \(x = 3\) with \(y = 0\). Wait, but also the interval \([-3, -2]\): The point \((-2.5, 0)\) is on \([-3, -2]\)? Wait \(-2.5\) is between -3 and -2. But the local minimum in \([-3, -2]\) is the lowest point, which is \((-1.56, -6)\) (but that's at \(x \approx -1.56\), which is in \([-2, 0]\)). Wait, no, let's check the y - value at the minimum of each interval:
- \([-3, -2]\): The function at \(x=-2.5\) is 0, but the local minimum (the lowest point) in this interval is the point \((-1.56, -6)\)? No, \(-1.56\) is greater than -2, so in \([-3, -2]\), the function goes from left (x=-3, some value) down to \((-2.5, 0)\)? Wait no, the graph: left side comes from top, goes down to \((-1.56, -6)\) (which is at x≈-1.56, between -2 and 0), then up to (0,0). Wait, maybe I messed up the intervals. Wait the interval \([2, 4]\): At x=3, y=0, and it's a local minimum (the bottom of the right valley). The interval \([-3, -2]\): The point (-2.5, 0) is on [-3, -2] (since -3 < -2.5 < -2). But is that a local minimum? Wait, a local minimum is a point where f(x) ≤ f(x + h) and f(x) ≤ f(x - h) for small h. At x=-2.5, if we move left (towards -3), the function was higher (since it came from the top left), and moving right (towards -2), it goes down to (-1.56, -6). So at x=-2.5, f(x)=0, but to the right, it's lower, so x=-2.5 is not a local minimum (it's a local maximum? No, wait the graph: left side, from x=-infinity, comes down, touches (-2.5, 0), then goes down to (-1.56, -6), then up to (0,0). So (-2.5, 0) is a local maximum? Wait, no, maybe I misinterpret the graph. Wait the graph has a root at (-2.5, 0), then a minimum at (-1.56, -6), then a root at (0,0), a peak at (1.2, 2.9), then a minimum at (3, 0), then goes up. So the local minimum at (3, 0) is in the interval [2, 4]. Wait, but let's check the y - value: at (3, 0), y=0. In [2, 4], the function goes from (1.2, 2.9) down to (3, 0) then up. So (3, 0) is a local minimum (the lowest point in [2, 4]) with value 0. Wait, but also the interval [-3, -2]: the point (-2.5, 0) is on it, but the local minimum there is not 0. The interval [2, 4]: the local minimum is (3, 0) with y=0. Wait, but let's check the options again. Wait the interval [2,4]: yes, because at x=3, y=0, and it's a local minimum (the function is decreasing before x=3 and increasing after, so x=3 is a local minimum). And the value is 0. The other intervals: [-3,-2] has a local minimum of -6, [-2,0] has local minimum -6, [1,2] has a peak then decreasing, no local minimum of 0. So the interval [2,4] has a local minimum of 0. Wait, but wait the point (3,0) is in [2,4], and it's a local minimum (since around x=3, the function is lower or equal? Wait, to the left of x=3 (in [2,3]), the function is decreasing from (1.2, 2.9) down to (3,0), and to the right (in [3,4]), it's increasing. So x=3 is a local minimum, with f(3)=0. So the interval [2,4] contains this local minimum with value 0.

Answer:

[2, 4] (the option D if we consider the options as A:[-3,-2], B:[-2,0], C:[1,2], D:[2,4])