**Mean Value Theorem hypothesis:** Function \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\).

For each function below, circle intervals when the conditions of the MVT are met. For non-circled intervals, provide a reason how the MVT hypothesis fails.

1. \( f(x) = x^2 \), on the intervals

- \( [4, 9] \) 😀
- \( [-4, 9] \) 😀
- \( [0, 9] \) 😀

2. \( f(x) = \sqrt{x} \), on the intervals

- \( [4, 9]\) 😀
- \( [-4, 9]\) fails: not continuous for \(x\lt0\)
- \( [0, 9]\) 😀

3. \( f(x) = \frac{1}{x} \), on the intervals

- \( [-10, 10]\) fails: not continuous at \(x=0\)
- \( [1, 10]\) 😀
- \( [-10, 0]\) fails: not continuous at \(x=0\)

4. \( f(x) = x^{2/3} \), on the intervals

- \( [-8, 8] \) fails: not differentiable at \(x=0\)
- \( [0, 8] \) 😀
- \( [1, 8] \) 😀

**Mean Value Theorem result:** Then there exists a number \(c \in (a,b)\) such that

\(f'(c) = \frac{f(b) - f(a)}{b - a}\)

Choose the first valid interval from the previous page.

- \(f(x) = x^2\), on the interval _____ \([4, 9]\)

\( \begin{aligned} m_{sec} = \frac{9^2 - 4^2}{9-4} = \frac{81 - 16}{5} = \frac{65}{5} = 13 \\ f'(c) = m_{sec} \\ 2c = 13 \\ c = \frac{13}{2} = 6.5 \end{aligned} \) - \(f(x) = \sqrt{x}\), on the interval _____ \([4, 9]\)

\( \begin{aligned} m_{sec} = \frac{\sqrt{9} - \sqrt{4}}{9-4} = \frac{3-2}{5} = \frac{1}{5} \\ f'(c) = m_{sec} \\ \frac{1}{2 \sqrt{c}} = \frac{1}{5} \\ 2 \sqrt{c} = 5 \\ \sqrt{c} = \frac{5}{2} \\ c = \frac{25}{4} \end{aligned} \) - \(f(x) = \frac{1}{x}\), on the interval _____ \([1, 10]\)

\( \begin{aligned} m_{sec} = \frac{\frac{1}{10} - \frac{1}{1}}{10-1} = \frac{\frac{-9}{10}}{9} = -\frac{1}{10} \\ f'(c) = m_{sec} \\ -\frac{1}{c^2} = -\frac{1}{10} \\ c^2 = 10 \\ c = \pm\sqrt{10} \\ c = \sqrt{10} \end{aligned} \) - \(f(x) = x^{2/3}\), on the interval _____ \([1, 8]\)

\( \begin{aligned} m_{sec} = \frac{8^{2/3} - 0^{2/3}}{8 - 0} = \frac{4 - 0}{8} = \frac{1}{2} \\ f'(c) = m_{sec} \\ \frac{2}{3 c^{1/3}} = \frac{1}{2} \\ \frac{4}{3} = c^{1/3} \\ \frac{64}{27} = c \end{aligned} \)

Next, find the number \(c\) satisfying the conclusion of the MVT for each of the above functions. Hints:

- Calculate the slope of the secant line between the endpoints, \(m_{sec} = \frac{ f(b)-f(a) }{ b-a }\)
- Solve for number \(c\) so that \(f'(c) = m_{sec}\) where \(c\) is in the interval \((a, b)\)

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**Application:** A trucker was given a speeding ticket at a toll booth after handing in her toll booth stub. The stub indicated she had covered 149 miles in 2 hours on a toll road with a speed limit of 65 mph. Was this speeding ticket justified?

\( m_{sec} = \frac{149}{2} = 74.5 \)

Driving is continuous and instantaneous velocities exist (e.g., the derivative exists). Therefore at some point the trucker must have had an instantaneous velocity of 74.5.