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.

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. $h'(3) = \frac{1}{2}$

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

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.