Mean Value Theorem (MVT)

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

  1. \( [4, 9] \)
  2. \( [-4, 9] \)
  3. \( [0, 9] \)

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

  1. \( [4, 9]\)
  2. \( [-4, 9]\)
  3. \( [0, 9]\)

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

  1. \( [-10, 10]\)
  2. \( [1, 10]\)
  3. \( [-10, 0]\)

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

  1. \( [-8, 8] \)
  2. \( [0, 8] \)
  3. \( [1, 8] \)

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 …

  1. Calculate the slope of the secant line between the endpoints, that is, \(m_{sec} = \frac{ f(b)-f(a) }{ b-a }\)
  2. Calculate \( f'(x) \).
  3. 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.

Practice computing "c"

  1. \(f(x) = x^2\), on the interval \([4, 9]\)
  2. \(f(x) = \sqrt{x}\), on the interval \([4, 9]\)
  3. \(f(x) = \frac{1}{x}\), on the interval \([1, 10]\)
  4. \(f(x) = x^{2/3}\), on the interval \([0, 8]\)

Application: Speeding tickets

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?