We are asked to find $H'(5)$, where $H(x) = 4f(x) + 2g(x)$, based on the graphs of $f(x)$ and $g(x)$.

To do this, we need the following steps:

1. Find the derivatives $f'(x)$ and $g'(x)$ at $x = 5$ from the graphs, as these derivatives correspond to the slopes of the lines $y = f(x)$ and $y = g(x)$ at $x = 5$.
2. Then use the derivative rule for linear combinations of functions: $H'(x) = 4f'(x) + 2g'(x)$
3. Evaluate this at $x = 5$ to find $H'(5)$.

Step 1: Find $f'(5)$ and $g'(5)$ from the graphs

• From the graph, $f(x)$ appears to be a straight line. The slope of a straight line is constant, so we can calculate $f'(x)$ by finding the slope.
• Similarly, we will calculate the slope for $g(x)$.

Calculating Slopes:

1. For $f(x)$ (blue line): The line passes through points (0, 1) and (5, 3). The slope is given by: $f'(x) = \frac{3 - 1}{5 - 0} = \frac{2}{5}$ So, $f'(5) = \frac{2}{5}$.

2. For $g(x)$ (red line): The line passes through points (0, 10) and (5, 0). The slope is given by: $g'(x) = \frac{0 - 10}{5 - 0} = \frac{-10}{5} = -2$ So, $g'(5) = -2$.

Step 2: Use the derivative rule for $H'(x)$

$H'(x) = 4f'(x) + 2g'(x)$ Substitute $f'(5) = \frac{2}{5}$ and $g'(5) = -2$: $H'(5) = 4\left(\frac{2}{5}\right) + 2(-2) = \frac{8}{5} - 4 = \frac{8}{5} - \frac{20}{5} = \frac{-12}{5}$

Thus, $H'(5) = \frac{-12}{5}$.

Would you like further details on this solution or have any other questions?

Here are 5 related questions:

1. How do we find the derivative of a function graphically?
2. What is the significance of linearity in determining derivatives?
3. How would $H'(x)$ change if the coefficients of $f(x)$ and $g(x)$ were different?
4. How can we approximate slopes if the graphs aren't straight lines?
5. How do we apply the chain rule to more complex functions?

Tip: For straight lines, the derivative is simply the slope, which makes calculating derivatives visually intuitive.