The Theorem
If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one c in (a,b) such that:
f'(c) = [f(b) − f(a)] / (b − a)
Example 1 — Basic Verification
📋 Verify MVT for f(x) = x² on [1,4], then find c
Check conditionsx² is continuous and differentiable everywhere ✓
Average rate[f(4)−f(1)]/(4−1) = (16−1)/3 = 5
Set f'(c)=52c = 5 → c = 5/2 = 2.5
Check2.5 ∈ (1,4) ✓
Example 2 — Cubic Function
📋 f(x) = x³ − x on [0,2], find all c
Average rate[f(2)−f(0)]/(2−0) = (6−0)/2 = 3
f'(x) = 33x²−1 = 3 → 3x² = 4 → x = ±√(4/3) = ±2/√3
Valid cOnly c = 2/√3 ≈ 1.155 ∈ (0,2) ✓
Rolle's Theorem (Special Case)
When f(a) = f(b), the MVT becomes Rolle's Theorem: there exists c where f'(c) = 0.
📋 Verify Rolle's for f(x) = x²−4x+3 on [1,3]
Checkf(1) = 1−4+3 = 0, f(3) = 9−12+3 = 0 ✓
Set f'(c)=02c−4 = 0 → c = 2 ∈ (1,3) ✓
Example 4 — Trig Function
📋 f(x) = sin(x) on [0, π], find c
Average rate[sin(π)−sin(0)]/(π−0) = 0/π = 0
Set f'(c)=0cos(c) = 0 → c = π/2
Checkπ/2 ∈ (0,π) ✓
Example 5 — Not Differentiable (MVT Does Not Apply)
📋 f(x) = |x| on [−1, 1] — does MVT apply?
Conditionsf is continuous on [−1,1] ✓
Differentiable?f is NOT differentiable at x=0 (corner) ✗
ConclusionMVT does NOT apply. Indeed, [f(1)−f(−1)]/(1−(−1)) = 0/2 = 0 but f'(x) is never 0 on (−1,1).
Example 6 — Application: Speed
📋 A car travels 150 miles in 2.5 hours. Was the car ever going exactly 60 mph?
SetupLet p(t) = position. Average rate = 150/2.5 = 60 mph
MVTIf p is continuous and differentiable (reasonable for a car), then there exists c where p'(c) = 60 mph
AnswerYes — the car's speedometer must have read exactly 60 mph at some moment
Example 7 — Proving Inequality
📋 Prove sin(x) ≤ x for all x ≥ 0
StrategyApply MVT to f(t) = sin(t) on [0, x] for any x > 0
MVT gives∃c ∈ (0,x) where sin(x)−sin(0) = cos(c)·(x−0)
Since|cos(c)| ≤ 1, we get |sin(x)| ≤ |x|, so sin(x) ≤ x for x ≥ 0 ✓
Example 8 — Finding Number of Roots
📋 Show f(x) = x³ + 2x − 1 has exactly one real root
Existencef(0) = −1 < 0, f(1) = 2 > 0. By IVT, at least one root in (0,1) ✓
Uniquenessf'(x) = 3x²+2 > 0 always → f is strictly increasing
ConclusionA strictly increasing function can cross zero at most once → exactly one root ✓
Frequently Asked Questions
Can there be more than one c in the MVT?
Yes — the theorem guarantees "at least one" c. A function like sin(x) over a large interval will have many points where the tangent is parallel to the secant. The theorem doesn't tell you how many — just that one exists.
What is the difference between MVT and IVT?
The Intermediate Value Theorem concerns the values a continuous function takes (it must hit every value between f(a) and f(b)). The Mean Value Theorem concerns the derivative — it guarantees a point where the instantaneous rate equals the average rate. They're different theorems requiring different conditions.