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To solve this problem, we need to calculate the derivatives of the product \( h(x) = f(x) \cdot g(x) \) at specific points: \( x = 1 \), \( x = 2 \), and \( x = 3 \). 

We will apply the **product rule** for differentiation:
\[
h'(x) = f'(x)g(x) + f(x)g'(x)
\]

### Steps:

1. **Estimate \( f(x) \) and \( g(x) \) values**: We will estimate the values of the functions \( f(x) \) and \( g(x) \) by observing their respective graphs at \( x = 1 \), \( x = 2 \), and \( x = 3 \).
   
2. **Estimate \( f'(x) \) and \( g'(x) \) derivatives**: From the graphs, we will estimate the slopes of the functions at \( x = 1 \), \( x = 2 \), and \( x = 3 \). For \( x = 2 \), the derivative does not exist for \( f(x) \) because of the sharp corner.

3. **Substitute the values** into the product rule and calculate the derivative.

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### Part A: \( h'(1) \)

- From the graph:
  - \( f(1) = 1 \), \( g(1) = 0 \) (from their respective points on the graph).
  - The slopes at \( x = 1 \):
    - \( f'(1) \) (slope of \( f(x) \)) is \( \frac{2-0}{2-0} = 1 \).
    - \( g'(1) \) (slope of \( g(x) \)) is \( \frac{0 - (-1)}{2 - 0} = \frac{1}{2} \).
  
  Now applying the product rule:
  \[
  h'(1) = f'(1)g(1) + f(1)g'(1) = 1(0) + 1\left(\frac{1}{2}\right) = \frac{1}{2}
  \]

### Part B: \( h'(2) \)

- \( f(x) \) has a sharp corner at \( x = 2 \), meaning the derivative \( f'(2) \) **does not exist**. Therefore:
  \[
  h'(2) = \text{DNE}
  \]

### Part C: \( h'(3) \)

- From the graph:
  - \( f(3) = 1 \), \( g(3) = -1 \).
  - The slopes at \( x = 3 \):
    - \( f'(3) \) (slope of \( f(x) \)) is \( \frac{0 - 2}{4 - 2} = -1 \).
    - \( g'(3) \) (slope of \( g(x) \)) is \( \frac{-1 - 0}{4 - 2} = -\frac{1}{2} \).
  
  Now applying the product rule:
  \[
  h'(3) = f'(3)g(3) + f(3)g'(3) = (-1)(-1) + 1\left(-\frac{1}{2}\right) = 1 - \frac{1}{2} = \frac{1}{2}
  \]

### Final Answers:

- **A**. \( h'(1) = \frac{1}{2} \)
- **B**. \( h'(2) = \text{DNE} \)
- **C**. \( h'(3) = \frac{1}{2} \)

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Do you have any questions or would you like further details?

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Here are 5 related questions to deepen your understanding:

1. How do we calculate the derivative of a product of two functions using the product rule?
2. Why does a sharp corner cause the derivative to not exist at a point?
3. What is the difference between a sharp corner and a discontinuity when determining derivatives?
4. How can we estimate the slope of a graph visually when calculating derivatives?
5. Can the derivative of a product exist if one of the factors has a discontinuity?

**Tip**: Always be cautious when identifying points where the derivative does not exist. Corners, cusps, and vertical tangents are common causes for this.

Use the figure below to estimate the indicated derivatives, or state that they do not exist. Let h(x) = f(x) * g(x). Find h'(1), h'(2), and h'(3).

Solve a calculus problem involving the estimation of derivatives using the product rule and analyzing graphs. This includes the functions f(x) and g(x), with a sharp corner in f(x) at x=2, causing the derivative to not exist there.

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