Let's go through the problems step by step from this worksheet.

a. The function $k(x)$ is defined by $k(x) = f(x)g(x)$. Find $k'(0)$. (3 points)

To find $k'(x)$, we use the product rule: $k'(x) = f'(x)g(x) + f(x)g'(x).$

We are asked to find $k'(0)$, so: $k'(0) = f'(0)g(0) + f(0)g'(0).$

From the graph of $f(x)$, we can estimate $f(0)$, $f'(0)$, and then we need to know $g(0)$ and $g'(0)$.

Based on the graph and given functions, you can:

1. Evaluate $f(0)$ from the graph directly.
2. Use $g(x) = \sqrt{x^2 - x + 3}$ to find $g(0)$ and $g'(0)$.

Once you have these values, you can plug them into the product rule expression to get $k'(0)$.

b. The function $m(x)$ is defined by $m(x) = \frac{f(x)}{2g(x)}$. Find $m'(5)$. (3 points)

To differentiate this function, we'll apply the quotient rule: $m'(x) = \frac{f'(x) \cdot 2g(x) - f(x) \cdot 2g'(x)}{(2g(x))^2}.$

We are asked to find $m'(5)$, so: $m'(5) = \frac{f'(5) \cdot 2g(5) - f(5) \cdot 2g'(5)}{(2g(5))^2}.$

Steps:

1. Find $f(5)$ and $f'(5)$ from the graph.
2. Use the expression for $g(x) = \sqrt{x^2 - x + 3}$ to find $g(5)$ and $g'(5)$.

Plug these values into the quotient rule formula to get $m'(5)$.

c. Find the value of $x$ for $-1 < x < 2$ such that $f'(x) = h'(x)$. (2 points)

We are looking for the value of $x$ where the derivatives of $f(x)$ and $h(x)$ are equal.

Steps:

1. Use the graph to find $f'(x)$ in the interval $-1 < x < 2$.
2. Use the derivative of $h(x) = 5e^x - 9\sin(x)$, which is: $h'(x) = 5e^x - 9\cos(x).$
3. Set $f'(x) = h'(x)$ and solve for $x$ in the given interval.

Would you like to go deeper into any of these parts?

Here are five related follow-up questions:

1. How is the product rule applied to more complex combinations of functions?
2. Can you explain the steps in applying the quotient rule to rational functions?
3. How do we estimate derivative values using graphs?
4. How do we solve transcendental equations, like $f'(x) = h'(x)$, involving both exponentials and trigonometric functions?
5. What is the significance of understanding both local behavior (derivatives) and global behavior of functions?

Tip: When solving problems involving graphs, make sure to carefully analyze the slope (derivative) at key points to accurately find values like $f'(x)$.