Let's first focus on part (b) of the problem, which asks for the derivative of $v(x) = \frac{f(x)}{g(x)}$ at $x = 6$.

To find $v'(x)$, we will apply the quotient rule. The quotient rule states that:

$v'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

Now, we need to find the values for $f(6), f'(6), g(6),$ and $g'(6)$ using the graph.

1. For $f(6)$ and $g(6)$:

   • From the graph, at $x = 6$, $f(6) = 5$ and $g(6) = 5$.

2. For $f'(6)$ and $g'(6)$:

   • $f'(6)$ is the slope of the function $f(x)$ at $x = 6$. The line for $f(x)$ between $x = 4$ and $x = 6$ has a constant slope. The slope is: $f'(6) = \frac{f(6) - f(4)}{6 - 4} = \frac{5 - 2}{2} = \frac{3}{2}$
   • Similarly, $g'(6)$ is the slope of $g(x)$ at $x = 6$. The line for $g(x)$ between $x = 5$ and $x = 6$ has a constant slope. The slope is: $g'(6) = \frac{g(6) - g(5)}{6 - 5} = \frac{5 - 1}{1} = 4$

Now we can substitute these values into the quotient rule formula:

$v'(6) = \frac{f'(6)g(6) - f(6)g'(6)}{[g(6)]^2} = \frac{\left(\frac{3}{2} \cdot 5 \right) - (5 \cdot 4)}{5^2}$

Simplifying:

$v'(6) = \frac{\frac{15}{2} - 20}{25} = \frac{-\frac{25}{2}}{25} = -\frac{1}{2}$

Thus, $v'(6) = -\frac{1}{2}$.

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

Related Questions:

1. What is the quotient rule, and when is it applied?
2. How do you find the slope of a linear function from a graph?
3. Can the slope of a piecewise function change at different intervals?
4. How does the behavior of $u(x) = f(x)g(x)$ differ from $v(x) = \frac{f(x)}{g(x)}$?
5. Why is the denominator in the quotient rule squared?

Tip: When applying the quotient rule, always make sure to differentiate both the numerator and the denominator before substituting the values into the formula.