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 why 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
\(f'(c) = \frac{f(b) - f(a)}{b - a}\)
MVT is an existence theorem in that it tells us that a value called \(c\) exists, but not how to find it. We will go a step further and use algebra to find the number \(c\) satisfying the conclusion of the MVT for each of the situations below. To to this …
Calculate the slope of the secant line between the endpoints, that is, \(m_{sec} = \frac{ f(b)-f(a) }{ b-a }\)
Calculate \( f'(x) \).
Let \(f'(c) = m_{sec}\) then use algebra to solve for \(c\). Recall that \( c \in (a, b) \).
Note: We are using the first valid interval from the previous section.
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.