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Using the Mean Value Theorem AB & BC

The Mean Value Theorem requires $f$ to be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, and then guarantees at least one $c$ in $(a, b)$ with $f'(c) = \frac{f(b) - f(a)}{b - a}$. The two intervals are different on purpose: the conclusion is built from $f(a)$ and $f(b)$, so the endpoints must be reachable, while $c$ is always strictly interior, so the derivative at an endpoint is never consulted. A vertical tangent at an endpoint costs nothing, as $x^{1/3}$ on $[0, 8]$ shows; a corner one step inside is fatal, as $|x|$ on $[-1, 1]$ shows.

Two failures dominate. The hypotheses go unchecked, and the theorem is applied across a corner, a jump, or an asymptote, where it produces an answer that nothing supports. And the conclusion is overstated: it promises at least one $c$, not exactly one, not the midpoint, and it reports a value of $f'$ rather than of $f$. Rolle's Theorem is the same statement with the extra hypothesis $f(a) = f(b)$, which is the hypothesis most often skipped.

f(x) = x³ − 3x ON [−2, 2]. AVERAGE RATE = (2 − (−2)) / 4 = 1 THE DASHED SECANT HAS SLOPE 1. FIND EVERY TANGENT PARALLEL TO IT. c = −1.155 c = +1.155 (−2, −2) (2, 2) AT LEAST ONE c IS PROMISED. THIS CURVE HAS TWO, AND NEITHER IS THE MIDPOINT 0. HYPOTHESES ARE CHECKED ON THE CLOSED [−2, 2]. EVERY c LIVES IN THE OPEN (−2, 2).
Drawn to scale at 110 px per unit across and 45 px per unit up. Both green tangents really are parallel to the dashed secant.
WHERE THE FAILURE SITS DECIDES EVERYTHING. f(x) = |x| f(x) = 1/x f(x) = x^(1/3) ON [−1, 1] ON [−1, 2] ON [0, 8] CONTINUOUS? YES CONTINUOUS? NO CONTINUOUS? YES DIFFERENTIABLE? NO UNDEFINED AT x = 0 f′ FAILS AT x = 0 a CORNER AT x = 0, and 0 is INSIDE the but 0 is an ENDPOINT, INSIDE the interval closed interval not inside (0, 8) AVG RATE = 0, BUT AVG RATE = 1/2, BUT AVG RATE = 1/4, AND SLOPES ARE ±1: NO c f′ ≤ 0 ALWAYS: NO c c = 1.540 EXISTS CONTINUITY ON THE CLOSED INTERVAL, DIFFERENTIABILITY ON THE OPEN ONE. A POINT WHERE f′ FAILS COSTS NOTHING IF IT IS AN ENDPOINT, AND EVERYTHING IF IT IS NOT.
Two of these are rejected on sight by a careless reader and one is accepted on sight. Each time, the careless reading is wrong.

The work

3 ways in · any order
Lesson
Using the Mean Value Theorem ›

Separates the two Mean Value Theorem hypotheses and shows why they live on different intervals, works the counterexamples where continuity or differentiability fails inside, shows the endpoint case that looks fatal and is not, and pins down what the conclusion promises: at least one interior point, a value of the derivative, and nothing about the midpoint.

Skill check · 10 scenarios
Open lesson →
Diagnostic
10-item topic check ›

Ten items spanning the two failure modes of this topic: applying the theorem across a corner, a jump, or an asymptote without checking the hypotheses, and misreporting the conclusion as a unique point, as the midpoint, or as a statement about the function rather than its derivative.

Not yet available · 10 items
Coming soon
Targeted Practice
Drill a single misconception ›

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.

Take the diagnostic to identify your misconceptions
Diagnostic first