Mean Value Theorem Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

easy
Find the value of cc guaranteed by the MVT for f(x)=x3f(x) = x^3 on [0,2][0, 2].

Solution

  1. 1
    Average rate: f(2)−f(0)2−0=82=4\frac{f(2)-f(0)}{2-0} = \frac{8}{2} = 4.
  2. 2
    f′(x)=3x2f'(x) = 3x^2. Set 3c2=4⇒c2=43⇒c=23≈1.1553c^2 = 4 \Rightarrow c^2 = \frac{4}{3} \Rightarrow c = \frac{2}{\sqrt{3}} \approx 1.155.
  3. 3
    Check: c≈1.155∈(0,2)c \approx 1.155 \in (0,2). ✓

Answer

c=233≈1.155c = \frac{2\sqrt{3}}{3} \approx 1.155
Compute the average slope, set f′(c)f'(c) equal to it, and solve. The cc is not at the midpoint x=1x=1; its location depends on the function's curvature.

About Mean Value Theorem

If ff is continuous on [a,b][a, b] and differentiable on (a,b)(a, b), then there exists at least one point cc in (a,b)(a, b) where f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b) - f(a)}{b - a}

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More Mean Value Theorem Examples

Example 1 easy

Verify the MVT for [formula] on [formula] by finding the value of [formula].

Example 2 hard

Use the MVT to prove: if [formula] for all [formula] in [formula], then [formula] is constant on [fo

Example 4 medium

A car travels 120 miles in 2 hours. Explain why the MVT guarantees the car exceeded 60 mph at some i

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