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.