To solve for $h'(3)$ given $h(x) = f(x)g(x)$, we need to use the product rule for differentiation:

$h'(x) = f'(x)g(x) + f(x)g'(x)$

Our task is to find $h'(3)$, which means evaluating $h'(x)$ at $x = 3$.

Step 1: Identify $f(x)$ and $g(x)$ at $x = 3$

From the graph:

• $f(3)$ appears to be around $4$.
• $g(3)$ appears to be around $2$.

Step 2: Determine $f'(3)$ and $g'(3)$

For $f(x)$:

The function $f(x)$ is linear, so it has a constant slope. Observing the line, $f(x)$ has a negative slope. We can calculate the slope by choosing two points on the line:

• At $x = 0$, $f(x) = 7$.
• At $x = 6$, $f(x) = 0$.

The slope of $f(x)$, or $f'(x)$, is: $f'(x) = \frac{0 - 7}{6 - 0} = -\frac{7}{6}$

Thus, $f'(3) = -\frac{7}{6}$.

For $g(x)$:

The function $g(x)$ is piecewise linear. At $x = 3$, we need the slope of $g(x)$ in the interval around $x = 3$. Observing the line segments, between $x = 2$ and $x = 4$:

• At $x = 2$, $g(x) = 1$.
• At $x = 4$, $g(x) = 3$.

The slope of $g(x)$, or $g'(x)$, in this interval is: $g'(x) = \frac{3 - 1}{4 - 2} = 1$

Thus, $g'(3) = 1$.

Step 3: Apply the Product Rule

Now we substitute these values into the product rule formula: $h'(3) = f'(3)g(3) + f(3)g'(3)$

Substituting the values: $h'(3) = \left(-\frac{7}{6}\right)(2) + (4)(1)$

Calculating each term:

1. $f'(3)g(3) = -\frac{7}{6} \times 2 = -\frac{7}{3}$
2. $f(3)g'(3) = 4 \times 1 = 4$

Adding these: $h'(3) = -\frac{7}{3} + 4 = -\frac{7}{3} + \frac{12}{3} = \frac{5}{3}$

Final Answer

$h'(3) = \frac{5}{3}$

Would you like further details or have any questions?

1. How is the product rule used in this problem?
2. Why is it necessary to calculate the slopes $f'(3)$ and $g'(3)$?
3. How do we interpret a piecewise function in derivative calculations?
4. What are the implications if $g(x)$ were not piecewise?
5. Could this problem be solved using another method?

Tip: When dealing with piecewise functions, always check the interval around the point of interest to determine the correct slope.