Function $h$ is graphed. Select all the intervals where $h^{\prime}(x)0$ and $h^{\prime \prime}(x)

Q: Function $h$ is graphed. Select all the intervals where $h^{\prime}(x)>0$ and $h^{\prime \prime}(x)<0$. Choose all answers that apply: A $-4.5<x<-3$ B $-3<x<-1$ (C) $-1<x<1$ D None of the above

∻Solution by Steps∻ ‖ step 1 ⋮ To solve for $h'(x) > 0$ and $h''(x) 0$ means the function is increasing, and $h''(x) 0$ and $h''(x) 0$ and $h''(x) 0$ means the function is increasing, and $h''(x) 0$ and $h''(x) 0$ and $h''(x) 0$ and $h''(x) 0$ means the function is increasing, and $h''(x) 0$ but $h''(x) \geq 0$ in this interval. ‖ ‖ step 4 ⋮ Thus, $h'(x) > 0$ and $h''(x) 0$ and $h''(x) 0$ and $h''(x)

Question

Function $h$ is graphed.

Select all the intervals where $h^{\prime}(x)>0$ and $h^{\prime \prime}(x)<0$.
Choose all answers that apply:
A $-4.5<x<-3$
B $-3<x<-1$
(C) $-1<x<1$

D None of the above

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To solve for $h'(x) > 0$ and $h''(x) < 0$ in the interval $-4.5 < x < -3$, we analyze the graph of $h(x)$. ‖ ‖ step 2 ⋮ From the graph, $h'(x) > 0$ means the function is increasing, and $h''(x) < 0$ means the function is concave down. ‖ ‖ step 3 ⋮ In the interval $-4.5 < x < -3$, the function $h(x)$ is decreasing, not increasing. Therefore, $h'(x) \leq 0$ in this interval. ‖ ‖ step 4 ⋮ Thus, $h'(x) > 0$ and $h''(x) < 0$ does not hold for $-4.5 < x < -3$. ‖ ‖ step 1 ⋮ To solve for $h'(x) > 0$ and $h''(x) < 0$ in the interval $-3 < x < -1$, we analyze the graph of $h(x)$. ‖ ‖ step 2 ⋮ From the graph, $h'(x) > 0$ means the function is increasing, and $h''(x) < 0$ means the function is concave down. ‖ ‖ step 3 ⋮ In the interval $-3 < x < -1$, the function $h(x)$ is increasing and concave down. Therefore, $h'(x) > 0$ and $h''(x) < 0$ in this interval. ‖ ‖ step 4 ⋮ Thus, $h'(x) > 0$ and $h''(x) < 0$ holds for $-3 < x < -1$. ‖ ‖ step 1 ⋮ To solve for $h'(x) > 0$ and $h''(x) < 0$ in the interval $-1 < x < 1$, we analyze the graph of $h(x)$. ‖ ‖ step 2 ⋮ From the graph, $h'(x) > 0$ means the function is increasing, and $h''(x) < 0$ means the function is concave down. ‖ ‖ step 3 ⋮ In the interval $-1 < x < 1$, the function $h(x)$ is increasing and concave up. Therefore, $h'(x) > 0$ but $h''(x) \geq 0$ in this interval. ‖ ‖ step 4 ⋮ Thus, $h'(x) > 0$ and $h''(x) < 0$ does not hold for $-1 < x < 1$. ‖ ∻[question number] Answer∻ ***B*** ∻Key Concept∻ ⚹Concavity and Increasing/Decreasing Intervals⚹ ∻Explanation∻ ⚹To determine where $h'(x) > 0$ and $h''(x) < 0$, we need to find intervals where the function is increasing and concave down.⚹◊What is the relationship between $h'(x)$ and $h''(x)$ in terms of intervals where $h'(x)>0$ and $h''(x)<0$?⍭ Generate me a similar question◊

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