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.