Calculus I › Applications of derivatives › full formula sheet
The Mean Value Theorem
Averaged 60 mph over two hours? Then at some instant your speedometer read exactly 60.
Requirements: f continuous on [a, b] and differentiable on (a, b). The theorem promises a c — it does not hand you its value.
Before this lesson: Rolle’s theorem · Definition of the derivative
Where it comes from
The problem: you drive 120 miles in 2 hours — average speed 60 mph. The naive claim, “you drove 60 the whole time,” is obviously false (traffic lights exist). The true claim is subtler: at some instant, your speedometer read exactly 60. You can’t average 60 without at some moment being at 60 — to get from below-average to above-average speed, you must cross 60 itself.
Geometrically: the secant line between (a, f(a)) and (b, f(b)) has some slope; roll a tangent line along the curve and at some point it will sit parallel to that secant. That point is c.
But the hypotheses are load-bearing, not decoration. Watch the conclusion fail when differentiability is dropped — f(x) = |x| on [−1, 1]:
Continuity matters too: take f(x) = x on [0, 1) with f(1) = 0. Then f(0) = f(1) = 0, average slope 0, yet f′(c) = 1 everywhere on (0, 1) — no flat tangent, because the jump at x = 1 broke continuity. Check the hypotheses first, always.
Before reading on: if the secant from (a, f(a)) to (b, f(b)) were tilted flat, what theorem would the result be? (It has its own page.)
Derivation
The Mean Value Theorem is Rolle’s theorem, tilted. Rolle needs equal endpoint values; our f doesn’t have them — so we subtract the secant line to force them, apply Rolle, and tilt back.
Why this trick works: subtracting the secant doesn’t change any slopes — it just slides the picture so the secant becomes horizontal. A horizontal secant is exactly Rolle’s setup. The whole proof is one clever subtraction.
Before reading on: f(x) = |x| on [−1, 1] has average slope 0. Predict whether some c has f′(c) = 0 — and what that tells you about the hypotheses.
How to use it
The procedure, every time:
- Verify the hypotheses. f continuous on the closed interval [a, b], differentiable on the open interval (a, b). If either fails, stop — the conclusion may be false (|x| says hello).
- Compute the average rate m = [f(b) − f(a)] / (b − a).
- Set f′(c) = m and solve for c. Keep only solutions with a < c < b — a c outside the interval doesn’t count.
- State the conclusion: this c satisfies the theorem. (There may be more than one.)
Existence vs. location
The theorem guarantees a c; it never computes it. In exercises you find c yourself by solving f′(c) = m — the theorem is what assures you the equation has a solution in (a, b).
The bounding superpower
Rearranged, MVT says f(b) − f(a) = f′(c)(b − a). If |f′| ≤ M everywhere, then |f(b) − f(a)| ≤ M(b − a) — a speed limit on how fast f can change. This one line powers error estimates across analysis.
Worked examples
Four problems, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — the basic move: f(x) = x² on [1, 3]
- Hypotheses. x² is a polynomial — continuous on [1, 3], differentiable on (1, 3). ✓
- Average rate. m = (9 − 1) / (3 − 1) = 8/2 = 4.
- Solve f′(c) = m. f′(x) = 2x, so 2c = 4 → c = 2.
- Check the interval. 1 < 2 < 3 ✓ — c = 2 works. (At x = 2 the tangent has slope 4, parallel to the secant.)
Your turn: f(x) = x² on [0, 2] — find c.
Answer: c = 1.
m = (4 − 0)/2 = 2; 2c = 2 gives c = 1, and 0 < 1 < 2 ✓.
Example 2 — with a root: f(x) = √x on [1, 4]
- Hypotheses. √x is continuous on [1, 4] and differentiable on (1, 4). ✓
- Average rate. m = (2 − 1) / (4 − 1) = 1/3.
- Solve f′(c) = m. f′(x) = 1/(2√x), so 1/(2√c) = 1/3 → 2√c = 3 → √c = 3/2 → c = 9/4 = 2.25.
- Check the interval. 1 < 2.25 < 4 ✓.
Your turn: f(x) = √x on [4, 9] — find c.
Answer: c = 25/4 = 6.25.
m = (3 − 2)/5 = 1/5; 1/(2√c) = 1/5 gives √c = 5/2, c = 25/4; 4 < 6.25 < 9 ✓.
Example 3 — the speeding ticket: 120 miles in 2 hours
- Set up. Let s(t) = position. s is continuous and differentiable (motion is smooth). a = 0, b = 2.
- Average rate. [s(2) − s(0)] / 2 = 120/2 = 60 mph.
- Conclude. MVT guarantees some c in (0, 2) with s′(c) = 60 — at some instant you were going exactly 60 mph. (Why does this convict speeders? If the speed limit were 55, that instant c is your ticket.)
Your turn: You drive 180 miles in 3 hours (limit 65 mph). Does MVT convict you?
Answer: No.
Average speed = 180/3 = 60 mph; MVT promises some instant at exactly 60 mph, but 60 < 65 — no proof of speeding.
Example 4 — when it breaks: f(x) = |x| on [−1, 1]
- Hypotheses? |x| is continuous on [−1, 1] ✓ — but not differentiable at 0, which lies in (−1, 1). ✗ Stop.
- What goes wrong. Average slope = (1 − 1)/2 = 0. But f′(c) is −1 for c < 0 and +1 for c > 0 — never 0. No c exists.
- The lesson. The theorem’s conclusion can fail when a hypothesis fails. Checking isn’t bureaucracy — it’s the difference between a true statement and a false one.
Your turn: Does MVT apply to f(x) = 1/x on [−1, 1]?
Answer: No — stop at the hypotheses.
f is undefined (hence not continuous) at 0, which lies in [−1, 1]. The theorem stays silent.
Memorization tips
- Say it: “average slope equals some instantaneous slope.” That sentence is the theorem.
- Hypotheses chant: “continuous on closed, differentiable on open.” Closed [a, b] for continuity, open (a, b) for the c.
- MVT is Rolle tilted: subtract the secant, get equal endpoints, apply Rolle, tilt back. One subtraction is the whole proof.
- Guarantee, not GPS: the theorem promises a c exists; it never tells you where. You solve f′(c) = m yourself.
- The |x| alarm: whenever you see a corner, cusp, or jump, hear an alarm — check the hypotheses before concluding anything.
- Bounding form: |f(b) − f(a)| ≤ M(b − a) when |f′| ≤ M. A speed limit for functions.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and the Mean Value Theorem is yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What does the Mean Value Theorem say in plain English?
If f is continuous on [a, b] and differentiable on (a, b), then at some instant c between a and b the instantaneous rate f′(c) exactly equals the average rate [f(b)−f(a)]/(b−a). Averaged 60 mph over 2 hours? Your speedometer read exactly 60 at some moment.
Why does the Mean Value Theorem need continuity and differentiability?
Without them the conclusion can fail. For f(x) = |x| on [−1, 1] the average slope is 0, but no c has f′(c) = 0 — the corner at 0 breaks differentiability, and with it the theorem.
How is the Mean Value Theorem related to Rolle’s theorem?
MVT is Rolle’s theorem tilted: subtract the secant line from f to get a helper function φ with φ(a) = φ(b) = 0, apply Rolle’s to get φ′(c) = 0, and that unpacks to f′(c) = average slope.
Does the Mean Value Theorem tell me what c is?
No — it guarantees that some c in (a, b) exists but gives no formula for it. In problems you find c by solving f′(c) = average slope yourself.
What is the Mean Value Theorem used for?
Existence arguments (some instant with a given rate), and bounds: if |f′| ≤ M on [a, b] then |f(b)−f(a)| ≤ M(b−a). It’s also the engine behind many deeper theorems.
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