The problem in your image asks to find $H'(3)$, where $H(x) = 2f(x) + 5g(x)$. The graphs of $f(x)$ and $g(x)$ are provided, and we are tasked with calculating $H'(x)$.

Given that $H(x) = 2f(x) + 5g(x)$, its derivative is:

$H'(x) = 2f'(x) + 5g'(x)$

To find $H'(3)$, we need the slopes of $f(x)$ and $g(x)$ at $x = 3$, which are the values of $f'(3)$ and $g'(3)$.

Looking at the graph:

• $f(x)$ (blue line) is linear, so its slope is constant. It appears to have a negative slope, and you can estimate the rise over run to determine $f'(x)$.
• $g(x)$ (red line) also has a constant slope, but it's positive. Again, you can estimate the rise over run for $g'(x)$.

Let's find the slopes:

• $f'(x)$ appears to have a slope of $-1$.
• $g'(x)$ appears to have a slope of $2$.

Substituting these values into the derivative:

$H'(3) = 2(-1) + 5(2) = -2 + 10 = 8$

Thus, $H'(3) = 8$.

Would you like more details or have any questions?

Here are 5 related questions:

1. How do we determine the derivative of a linear function from its graph?
2. How would $H(x)$ change if the coefficients of $f(x)$ and $g(x)$ were different?
3. What if $f(x)$ and $g(x)$ were non-linear functions? How would the process differ?
4. Can you use this approach to calculate $H'(x)$ at a different point?
5. How does the concept of slope relate to the derivative?

Tip: When graphing linear functions, the slope remains constant at all points, making it easier to calculate derivatives directly from the graph.