Mean Value Theorem Formula

Mean value theorem is if f is continuous on [a, b] and differentiable on (a, b), then there exists at least one point c in (a, b) where f'(c) = f(b).

The Formula

f′(c)=f(b)−f(a)b−afor some c∈(a,b)f'(c) = \frac{f(b) - f(a)}{b - a} \quad \text{for some } c \in (a, b)

When to use: If you drive 150 miles in 2 hours, your average speed is 75 mph. The MVT says at some instant during the trip, your speedometer read exactly 75 mph. The instantaneous rate must equal the average rate at least once.

Quick Example

f(x)=x2f(x) = x^2 on [1,3][1, 3]. Average rate: 9−13−1=4\frac{9-1}{3-1} = 4.
f′(x)=2x=4⇒x=2f'(x) = 2x = 4 \Rightarrow x = 2. Indeed, c=2∈(1,3)c = 2 \in (1, 3).
At x=2x = 2, the tangent line is parallel to the secant line through (1,1)(1, 1) and (3,9)(3, 9).

Notation

MVT. f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a} is the average rate of change (secant slope); f′(c)f'(c) is the instantaneous rate (tangent slope).

What This Formula Means

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}

If you drive 150 miles in 2 hours, your average speed is 75 mph. The MVT says at some instant during the trip, your speedometer read exactly 75 mph. The instantaneous rate must equal the average rate at least once.

Formal View

If ff is continuous on [a,b][a, b] and differentiable on (a,b)(a, b), then ∃ c∈(a,b)\exists\, c \in (a, b) such that f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b) - f(a)}{b - a}. Equivalently: f(b)−f(a)=f′(c)(b−a)f(b) - f(a) = f'(c)(b - a).

Worked Examples

Example 1

easy
Verify the MVT for f(x)=x2f(x) = x^2 on [1,3][1, 3] by finding the value of cc.

Answer

c=2∈(1,3)c = 2 \in (1, 3), confirming the MVT.

First step

1
Check hypotheses: ff is a polynomial, so continuous on [1,3][1,3] and differentiable on (1,3)(1,3). ✓

Full solution

  1. 2
    Average rate of change: f(3)−f(1)3−1=9−12=4\frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = 4.
  2. 3
    Find cc: f′(x)=2xf'(x) = 2x. Set 2c=4⇒c=22c = 4 \Rightarrow c = 2.
  3. 4
    Check: c=2∈(1,3)c = 2 \in (1, 3). ✓
The MVT guarantees a cc where the instantaneous rate equals the average rate. Here c=2c=2 is where the tangent to y=x2y = x^2 is parallel to the secant through (1,1)(1,1) and (3,9)(3,9).

Example 2

hard
Use the MVT to prove: if f′(x)=0f'(x) = 0 for all xx in (a,b)(a, b), then ff is constant on (a,b)(a, b).

Example 3

medium
Use MVT to prove ∣cos⁡a−cos⁡b∣≤∣a−b∣|\cos a-\cos b|\le|a-b| for all real a,ba,b.

Common Mistakes

  • Skipping the differentiability check - a corner or cusp on (a,b)(a,b) voids the theorem even with continuity.
  • Confusing it with IVT - MVT equates a SLOPE (derivative) to the average slope, not a function value.
  • Assuming cc is unique or at the midpoint - there is at least one cc, and it need not be in the middle.

Why This Formula Matters

It is the theoretical hinge of differential calculus: it proves that 'always increasing' follows from f′>0f'>0, justifies antiderivative uniqueness, and underlies curve-sketching. It also formalizes the everyday fact that your average speed must equal your speedometer reading at some moment. Recognizing it by "Is ff continuous on [a,b][a,b] and differentiable on (a,b)(a,b), with a guarantee sought that some cc matches the average slope?" — rather than by familiar numbers — is what lets a student tell it apart from intermediate value theorem and rolle's theorem and average rate of change in a mixed problem set.

Frequently Asked Questions

What is the Mean Value Theorem formula?

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}

How do you use the Mean Value Theorem formula?

If you drive 150 miles in 2 hours, your average speed is 75 mph. The MVT says at some instant during the trip, your speedometer read exactly 75 mph. The instantaneous rate must equal the average rate at least once.

What do the symbols mean in the Mean Value Theorem formula?

MVT. f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a} is the average rate of change (secant slope); f′(c)f'(c) is the instantaneous rate (tangent slope).

Why is the Mean Value Theorem formula important in Math?

It is the theoretical hinge of differential calculus: it proves that 'always increasing' follows from f′>0f'>0, justifies antiderivative uniqueness, and underlies curve-sketching. It also formalizes the everyday fact that your average speed must equal your speedometer reading at some moment. Recognizing it by "Is ff continuous on [a,b][a,b] and differentiable on (a,b)(a,b), with a guarantee sought that some cc matches the average slope?" — rather than by familiar numbers — is what lets a student tell it apart from intermediate value theorem and rolle's theorem and average rate of change in a mixed problem set.

What do students get wrong about Mean Value Theorem?

The procedure for mean value theorem is the easy part; the trap is skipping the differentiability check. Asking "Is ff continuous on [a,b][a,b] and differentiable on (a,b)(a,b), with a guarantee sought that some cc matches the average slope?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

What should I learn before the Mean Value Theorem formula?

Before studying the Mean Value Theorem formula, you should understand: derivative, limit, intermediate value theorem.

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