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For the following exercises, use the following figure to find the indicated derivatives, if they exist. Let h(x) = f(x) + g(x). Find a. h′(1), b. h′(3), and c. h′(4). 

To solve the given problem, we need to find the derivatives of the function \( h(x) = f(x) + g(x) \) at the specified points \( x = 1, 3, 4 \).

The graph contains two functions, \( f(x) \) (solid line) and \( g(x) \) (dashed line). We will use the concept of piecewise linear functions and the slopes of the linear segments to calculate the derivatives at \( x = 1, 3, 4 \).

### Step 1: Analyze the slopes of \( f(x) \) and \( g(x) \)
The derivative of a linear function is the slope of the segment at a specific point.

- **For \( f(x) \)**:
  - From \( x = 0 \) to \( x = 2 \), the slope is negative.
  - From \( x = 2 \) to \( x = 4 \), the slope is positive.
  
- **For \( g(x) \)**:
  - From \( x = 0 \) to \( x = 2 \), the slope is positive.
  - From \( x = 2 \) to \( x = 4 \), the function is constant (slope 0).

### Step 2: Find \( f'(x) \) and \( g'(x) \) at the specified points

#### a. \( h'(1) \):
- At \( x = 1 \), both \( f(x) \) and \( g(x) \) are linear. 
  - For \( f(x) \), the slope from \( x = 0 \) to \( x = 2 \) is: 
    \[
    \text{slope of } f(x) = \frac{f(2) - f(0)}{2 - 0} = \frac{2 - 4}{2} = -1
    \]
  - For \( g(x) \), the slope from \( x = 0 \) to \( x = 2 \) is:
    \[
    \text{slope of } g(x) = \frac{g(2) - g(0)}{2 - 0} = \frac{2 - 0}{2} = 1
    \]
  
Thus, \( h'(1) = f'(1) + g'(1) = (-1) + (1) = 0 \).

#### b. \( h'(3) \):
- At \( x = 3 \), both functions are linear again.
  - For \( f(x) \), the slope from \( x = 2 \) to \( x = 4 \) is:
    \[
    \text{slope of } f(x) = \frac{f(4) - f(2)}{4 - 2} = \frac{4 - 2}{2} = 1
    \]
  - For \( g(x) \), the function is constant from \( x = 2 \) to \( x = 4 \), so:
    \[
    \text{slope of } g(x) = 0
    \]

Thus, \( h'(3) = f'(3) + g'(3) = 1 + 0 = 1 \).

#### c. \( h'(4) \):
- At \( x = 4 \), \( f(x) \) is still linear with the same slope as in part (b):
  \[
  f'(4) = 1
  \]
- For \( g(x) \), the function is constant, so:
  \[
  g'(4) = 0
  \]

Thus, \( h'(4) = f'(4) + g'(4) = 1 + 0 = 1 \).

### Final Answers:
- \( h'(1) = 0 \)
- \( h'(3) = 1 \)
- \( h'(4) = 1 \)

Would you like further clarification or details on any part of this solution?

### Additional Questions:
1. How would the answer change if \( g(x) \) had a different slope from \( x = 2 \) to \( x = 4 \)?
2. Why does the derivative of \( g(x) \) become zero between \( x = 2 \) and \( x = 4 \)?
3. How do the slopes of \( f(x) \) and \( g(x) \) affect the overall derivative of \( h(x) \)?
4. Can the derivative exist at a point where there is a corner in \( f(x) \) or \( g(x) \)?
5. What is the interpretation of a zero derivative in terms of the behavior of the function at that point?

### Tip:
Remember that the derivative at a point is the slope of the tangent line. For piecewise linear functions like \( f(x) \) and \( g(x) \), the derivative is simply the slope of each linear segment unless there is a sharp corner, in which case the derivative does not exist at that point.

For the following exercises, use the following figure to find the indicated derivatives, if they exist. Let h(x) = f(x) + g(x). Find a. h′(1), b. h′(3), and c. h′(4).

This exercise involves finding the derivatives of h(x) = f(x) + g(x) based on a piecewise linear graph. Key concepts include slope calculations, the sum rule of derivatives, and piecewise function analysis. This problem is suitable for high school students in Grades 11-12.

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