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We are asked to find the values of \( h'(x) \) where \( h(x) = f(g(x)) \), using the given graphs of \( f(x) \) (black, with a sharp corner at \( x = 2 \)) and \( g(x) \) (blue). To calculate these derivatives, we'll need to apply the chain rule: 

\[
h'(x) = f'(g(x)) \cdot g'(x)
\]

Here’s how we can tackle each part:

### A. \( h'(1) \)
1. **Step 1: Evaluate \( g(1) \)**
   From the blue graph, at \( x = 1 \), \( g(1) = 0 \).
   
2. **Step 2: Find \( f'(g(1)) = f'(0) \)**
   The black graph shows that \( f(x) \) has a slope of 1 when \( x = 0 \), so \( f'(0) = 1 \).

3. **Step 3: Find \( g'(1) \)**
   The slope of \( g(x) \) (blue) at \( x = 1 \) appears to be constant and negative. From the graph, it looks like \( g(x) \) goes from \( (0,1) \) to \( (2,-1) \), which gives a slope:
   \[
   g'(x) = \frac{-1 - 1}{2 - 0} = -1
   \]

4. **Step 4: Multiply**
   \[
   h'(1) = f'(g(1)) \cdot g'(1) = 1 \cdot (-1) = -1
   \]

### B. \( h'(2) \)
1. **Step 1: Evaluate \( g(2) \)**
   From the blue graph, \( g(2) = 1 \).

2. **Step 2: Find \( f'(g(2)) = f'(1) \)**
   The black graph has a constant slope of 1 when \( x = 1 \), so \( f'(1) = 1 \).

3. **Step 3: Find \( g'(2) \)**
   The slope of \( g(x) \) (blue) at \( x = 2 \) is still constant and negative with \( g'(2) = -1 \).

4. **Step 4: Multiply**
   \[
   h'(2) = f'(g(2)) \cdot g'(2) = 1 \cdot (-1) = -1
   \]

### C. \( h'(3) \)
1. **Step 1: Evaluate \( g(3) \)**
   From the blue graph, \( g(3) = 2 \).

2. **Step 2: Find \( f'(g(3)) = f'(2) \)**
   The black graph shows a sharp corner at \( x = 2 \), so the derivative \( f'(2) \) does not exist (dne).

3. **Step 3: Conclusion**
   Since \( f'(g(3)) \) does not exist, \( h'(3) \) does not exist either.

Thus, the answers are:
- A. \( h'(1) = -1 \)
- B. \( h'(2) = -1 \)
- C. \( h'(3) = \text{dne} \)

Would you like more details or have any questions? Here are some related questions:

1. How do we interpret the sharp corner on a graph?
2. What is the chain rule, and when do we use it?
3. How do we find the slope of a line from a graph?
4. What happens to a composite function’s derivative at a non-differentiable point?
5. How can we tell when a function’s derivative does not exist?

**Tip:** Always check if a function has a sharp corner or discontinuity when calculating derivatives, as this can affect differentiability.

Let h(x) = f(g(x)). Find h'(1), h'(2), h'(3) using the graph of f(x) (black, sharp corner at x=2) and g(x) (blue).

This problem uses the chain rule to differentiate the composite function h(x) = f(g(x)). Based on the given graphs of f(x) and g(x), the values of h'(1), h'(2), and h'(3) are calculated. The solution involves using the slopes of the graphs and identifying non-differentiable points.

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