Function
\[h\] is graphed.
Function h is graphed. The x-axis goes from negative 5 to 5. The graph consists of a curve. The curve starts in quadrant 3, moves upward concave down to about (0, negative 10), continues upward concave up, and ends in quadrant 1.
\[\small{1}\]
\[\small{2}\]
\[\small{3}\]
\[\small{4}\]
\[\small{\llap{-}1}\]
\[\small{\llap{-}2}\]
\[\small{\llap{-}3}\]
\[\small{\llap{-}4}\]
\[\small{30}\]
\[\small{60}\]
\[\small{90}\]
\[\small{\llap{-}30}\]
\[\small{\llap{-}60}\]
\[\small{\llap{-}90}\]
\[y\]
\[x\]
\[\blueD{y=h(x)}\]
Select all the intervals where
\[h'(x)>0\] and
\[h''(x)>0\].

To solve this problem, we need to understand the conditions given:

1. **\( h'(x) > 0 \)**: This condition means that the function \( h(x) \) is increasing. In other words, the slope of the curve is positive.

2. **\( h''(x) > 0 \)**: This condition indicates that the function \( h(x) \) is concave up. The curve bends upwards like a cup.

### Analysis of the Curve
From the description:
- The curve starts in quadrant 3, indicating it begins below and to the left of the origin.
- It moves upward concave down to about \( (0, -10) \). This means initially, the slope is positive but decreasing, suggesting concave down behavior (like an upside-down cup).
- After reaching \( (0, -10) \), it continues upward concave up, suggesting the slope is positive and increasing, indicating concave up behavior (like a right-side-up cup).
- The curve ends in quadrant 1, indicating it finishes above and to the right of the origin.

### Identifying Intervals for \( h'(x) > 0 \) and \( h''(x) > 0 \)

- **Where \( h'(x) > 0 \)**: The function is increasing, meaning the curve is going upwards. This occurs from when it starts moving upwards in quadrant 3 to where it continues upwards through quadrant 1.

- **Where \( h''(x) > 0 \)**: The function is concave up, meaning the curve bends upwards. According to the description, this change happens after \( (0, -10) \), continuing upwards with a concave up shape.

The interval we are looking for where both conditions are true is when the curve is increasing and concave up.

### Conclusion
From the analysis, the curve is increasing and concave up in the interval from \( x = 0 \) onwards to the right until the curve ends in quadrant 1. Thus:

- The interval where both \( h'(x) > 0 \) and \( h''(x) > 0 \) is: **\( (0, 5] \)**.

This interval corresponds to the function \( h(x) \) increasing and having a positive concavity. 

Do you want more details or have any questions?

Here are some related questions you might consider:

1. How do you determine where a function is concave down?
2. What are the graphical interpretations of the first and second derivatives?
3. How can you find the points of inflection from a graph?
4. What does it mean for a function to be decreasing and concave up?
5. How do you use the first and second derivatives to sketch the shape of a graph?

**Tip:** To find intervals of concavity and where the function is increasing or decreasing, use the first and second derivative tests to analyze the behavior of the function over its domain.

Understand how to analyze function behavior using derivatives. Learn about increasing functions and concave up behavior through a detailed example. Suitable for advanced high school students.

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