Question

Math

Posted 5 months ago

```
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
```

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Answer from Sia

Posted 5 months ago

Solution by Steps

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.

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