We are tasked with finding $h'(1)$, where $h(x) = f(g(x))$. This is the composition of two functions: $g(x) = 2x + 1$ and the piecewise linear function $f(x)$ as shown in the graph.

Step 1: Differentiate $h(x) = f(g(x))$

By the chain rule, we know that: $h'(x) = f'(g(x)) \cdot g'(x)$ where $g'(x)$ is the derivative of $g(x)$, and $f'(x)$ is the derivative of $f(x)$.

Step 2: Calculate $g'(x)$

The function $g(x) = 2x + 1$ is linear, so: $g'(x) = 2$

Step 3: Evaluate $g(1)$

Substitute $x = 1$ into $g(x)$: $g(1) = 2(1) + 1 = 3$ Thus, $g(1) = 3$, so we need to evaluate $f'(3)$.

Step 4: Determine $f'(3)$

Now, let's look at the graph of $f(x)$. The graph consists of two line segments:

• From $(0, 0)$ to $(1, 4)$
• From $(1, 4)$ to $(4, -2)$

The slope of the segment from $(1, 4)$ to $(4, -2)$ is: $\text{slope of } f(x) \text{ on this segment} = \frac{-2 - 4}{4 - 1} = \frac{-6}{3} = -2$ Since $3$ lies on this second segment, we conclude that $f'(3) = -2$.

Step 5: Calculate $h'(1)$

Now apply the chain rule: $h'(1) = f'(g(1)) \cdot g'(1) = f'(3) \cdot 2 = (-2) \cdot 2 = -4$

Thus, the value of $h'(1)$ is $\boxed{-4}$.

Would you like more details on any part of the explanation?

Related questions:

1. What is the geometric meaning of $f'(x)$ in this context?
2. How does the chain rule work for function compositions?
3. What are other examples where a piecewise function is involved in differentiation?
4. Can we explore other values of $h'(x)$ for different points besides $x = 1$?
5. How does the slope of a line segment impact the derivative of a function?

Tip:

Always check the graph of a piecewise linear function carefully to identify the correct slope (derivative) for the relevant interval.