which statement is true about the graphed function? f(x)0 over the intervals (-∞, -0.7) and (0.76, 2.5). f(x)0 over the intervals (-0.7, 0.76) and (0.76, ∞).

Question

which statement is true about the graphed function?
f(x)<0 over the intervals (-∞, -0.7) and (0.76, 2.5).
f(x)>0 over the intervals (-∞, -0.7) and (0.76, 2.5).
f(x)<0 over the intervals (-0.7, 0.76) and (2.5, ∞).
f(x)>0 over the intervals (-0.7, 0.76) and (0.76, ∞).

Accepted Answer

# Explanation: ## Step1: Analyze the graph's sign regions The graph of the function $ F(x) $ crosses the x - axis at $ x=- 0.7$, $ x = 0.76$ and $ x=2.5$. To determine where $ F(x)>0$ or $ F(x)<0$, we look at the y - values (the sign of the function) in the intervals separated by these x - intercepts. - For the interval $(-\infty,-0.7)$: We can see from the graph that the function is below the x - axis (since as $ x$ approaches $-\infty$, the function goes down and between $ x =-\infty$ and $ x=-0.7$, the y - values are negative), so $ F(x)<0$ in $(-\infty,-0.7)$ is wrong for the "greater than 0" case and $ F(x)>0$ in $(-\infty,-0.7)$ is wrong. - For the interval $(-0.7,0.76)$: The graph is above the x - axis (the vertex $(0,2)$ is in this interval and the function is above the x - axis between $ x=-0.7$ and $ x = 0.76$), so $ F(x)>0$ here. - For the interval $(0.76,2.5)$: The graph is below the x - axis (the point $(1.9,-5.7)$ is in this interval and the function is below the x - axis between $ x = 0.76$ and $ x=2.5$), so $ F(x)<0$ here. - For the interval $(2.5,\infty)$: The graph is above the x - axis (as $ x$ increases beyond $ x = 2.5$, the function goes up), so $ F(x)>0$ here. Now let's check each option: - Option 1: $ F(x)<0$ over $(-\infty,-0.7)$ (wrong, since in $(-\infty,-0.7)$ the function is negative? Wait no, wait the first interval $(-\infty,-0.7)$: looking at the graph, when $ x < - 0.7$, the function is below the x - axis (y is negative), and in $(0.76,2.5)$, the function is below the x - axis (y is negative). Wait, maybe I made a mistake earlier. Let's re - evaluate. Wait, the x - intercepts are at $ x=-0.7$, $ x = 0.76$ and $ x = 2.5$. The function is a cubic - like function? Wait, the graph has a local maximum at $(0,2)$ and a local minimum at $(1.9,-5.7)$. Let's use the method of testing intervals: 1. Interval $(-\infty,-0.7)$: Pick a test point, say $ x=-1$. From the graph, when $ x=-1$, the function is below the x - axis (y is negative), so $ F(-1)<0$. 2. Interval $(-0.7,0.76)$: Pick $ x = 0$, $ F(0)=2>0$, so $ F(x)>0$ here. 3. Interval $(0.76,2.5)$: Pick $ x = 1$, from the graph, the point $(1.9,-5.7)$ is in this interval, so $ F(1)<0$ (since $ x = 1$ is in $(0.76,2.5)$). 4. Interval $(2.5,\infty)$: Pick $ x = 3$, the function is above the x - axis, so $ F(3)>0$. Now let's check the options: - Option 1: $ F(x)<0$ over $(-\infty,-0.7)$ (since at $ x=-1$, $ F(x)<0$) and $(0.76,2.5)$ (at $ x = 1$, $ F(x)<0$). Let's check the other options: - Option 2: $ F(x)>0$ over $(-\infty,-0.7)$ (wrong, because at $ x=-1$, $ F(x)<0$) and $(0.76,2.5)$ (wrong, because at $ x = 1$, $ F(x)<0$). - Option 3: $ F(x)<0$ over $(-0.7,0.76)$ (wrong, because in $(-0.7,0.76)$ the function is above the x - axis, $ F(x)>0$) and $(2.5,\infty)$ (wrong, because in $(2.5,\infty)$ the function is above the x - axis, $ F(x)>0$). - Option 4: $ F(x)>0$ over $(-0.7,0.76)$ (correct) and $(0.76,\infty)$ (wrong, because in $(0.76,2.5)$ the function is below the x - axis, $ F(x)<0$). Wait, going back to Option 1: $ F(x)<0$ over $(-\infty,-0.7)$ (since for $ x < - 0.7$, the function is below the x - axis) and $(0.76,2.5)$ (since for $ 0.760$. - For $ 0.762.5$: The graph is above the x - axis (y - values are positive), so $ F(x)>0$. So Option 1 says $ F(x)<0$ over $(-\infty,-0.7)$ and $(0.76,2.5)$, which matches our analysis. Option 2 says $ F(x)>0$ over $(-\infty,-0.7)$ (wrong, since $ F(x)<0$ there) and $(0.76,2.5)$ (wrong, since $ F(x)<0$ there). Option 3 says $ F(x)<0$ over $(-0.7,0.76)$ (wrong, $ F(x)>0$ there) and $(2.5,\infty)$ (wrong, $ F(x)>0$ there). Option 4 says $ F(x)>0$ over $(-0.7,0.76)$ (correct) and $(0.76,\infty)$ (wrong, since $ F(x)<0$ in $(0.76,2.5)$). So the correct option is the first one: $ F(x)<0$ over the intervals $(-\infty,-0.7)$ and $(0.76,2.5)$. ## Step2: Verify each option - Option 1: - Interval $(-\infty,-0.7)$: As $ x$ approaches $-\infty$ and between $ x =-\infty$ and $ x=-0.7$, the graph is below the x - axis, so $ F(x)<0$. - Interval $(0.76,2.5)$: The graph is below the x - axis (the point $(1.9,-5.7)$ is in this interval), so $ F(x)<0$. This option is correct. - Option 2: - Interval $(-\infty,-0.7)$: $ F(x)<0$ here, not $>0$, so this option is wrong. - Interval $(0.76,2.5)$: $ F(x)<0$ here, not $>0$, so this option is wrong. - Option 3: - Interval $(-0.7,0.76)$: $ F(x)>0$ here (the vertex $(0,2)$ is in this interval), not $<0$, so this option is wrong. - Interval $(2.5,\infty)$: $ F(x)>0$ here, not $<0$, so this option is wrong. - Option 4: - Interval $(0.76,\infty)$: In $(0.76,2.5)$, $ F(x)<0$, so this option is wrong. # Answer: A. $ F(x) < 0 $ over the intervals $ (-\infty, -0.7) $ and $ (0.76, 2.5) $

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Step1: Analyze the graph's sign regions

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