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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.

If f and g are the functions whose graphs are shown, let u(x) = f(x)g(x) and v(x) = f(x)/g(x). (a) Find u'(1). (b) Find v'(6).

This problem demonstrates how to apply the product rule and quotient rule to find the derivatives of functions at specific points, with a focus on u'(1) and v'(6). It uses graphical interpretation of functions f(x) and g(x) and involves calculating slopes and applying differentiation formulas.

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