The problem asks to find $H'(3)$, where $H(x) = 5f(x) + g(x)$, and the graphs of $f(x)$ and $g(x)$ are provided.

We need to compute the derivative of $H(x)$, which is:

$H'(x) = 5f'(x) + g'(x)$

To find $H'(3)$, we must evaluate both $f'(3)$ and $g'(3)$. These derivatives represent the slopes of the tangent lines to the graphs of $f(x)$ and $g(x)$ at $x = 3$.

Step 1: Determine $f'(x)$ and $g'(x)$ from the graphs

• $f(x)$ appears as a linear function with a positive slope.
• $g(x)$ appears as a linear function with a negative slope.

Since both functions are linear, their slopes (derivatives) are constant. We can calculate the slopes from the graphs.

For $f(x)$:

Looking at the graph, $f(x)$ rises from $y = 0$ at $x = 0$ to $y = 6$ at $x = 6$. Thus, the slope $f'(x)$ is:

$f'(x) = \frac{6 - 0}{6 - 0} = 1$

For $g(x)$:

$g(x)$ falls from $y = 8$ at $x = 0$ to $y = 0$ at $x = 8$. Thus, the slope $g'(x)$ is:

$g'(x) = \frac{0 - 8}{8 - 0} = -1$

Step 2: Find $H'(3)$

Since the derivatives are constant for these linear functions, we have:

$f'(3) = 1 \quad \text{and} \quad g'(3) = -1$

Now, substitute into the expression for $H'(x)$:

$H'(3) = 5f'(3) + g'(3) = 5(1) + (-1) = 5 - 1 = 4$

Thus, $H'(3) = 4$.

Would you like further details on any step?

Related Questions:

1. How do you find the slope of a line using points on a graph?
2. Why is the derivative of a linear function constant?
3. How does the chain rule apply in more complex cases involving product or quotient of functions?
4. Can you generalize the result if $f(x)$ and $g(x)$ were nonlinear functions?
5. What is the geometric interpretation of the derivative of a function at a given point?

Tip:

For linear functions, the slope is constant and can be determined by selecting any two points on the line!