Mean value theorem

What you are looking at

The yellow chord $AB$ has gradient $\lambda = \dfrac{f(\beta) - f(\alpha)}{\beta - \alpha}$, the average rate of change. The theorem promises a $\xi$ in $(\alpha, \beta)$ where the tangent has the same gradient; the green lines are those tangents, and the working solves $f'(\xi) = \lambda$. On $f(x) = \dfrac{x^3}{3} - x$ over the starting interval $[-2, 1.5]$, $\lambda = \dfrac{1}{12}$ and there are two such points, $\xi = \pm\sqrt{\dfrac{13}{12}}$.

The conditions are Rolle’s without the equal heights. $\sqrt{x}$ on $[0, 4]$ meets them — it is continuous at $0$ and only has to be differentiable on the open interval — so $\xi$ exists: $\dfrac{1}{2\sqrt{\xi}} = \dfrac{1}{2}$ gives $\xi = 1$. $|x|$ on $[-1, 2]$ fails: it has a corner at $0$, and no tangent has the chord’s gradient $\dfrac{1}{3}$.

Try this

1. 1Drag $P$ until its tangent lines up with the chord, and compare $f'(x_P)$ in the working with $\lambda$.
2. 2Move $A$ and $B$: the chord’s gradient changes and so do the points $\xi$ — but on a smooth curve there is always at least one.
3. 3Choose $\sqrt{x}$ on $[0, 4]$: the tangent at $0$ is vertical, yet $\xi = 1$ exists, because differentiability is only needed on $(0, 4)$.
4. 4Choose $|x|$: the chord from $x = -1$ to $x = 2$ has gradient $\dfrac{1}{3}$, and every tangent has gradient $\pm 1$ — there is no $\xi$.

Mistakes students make

• Requiring differentiability at the endpoints: only continuity is needed there.
• Forgetting continuity on the closed interval: $f(x) = x$ on $[0, 1)$ with $f(1) = 0$ is differentiable on $(0, 1)$, but $\lambda = 0$ while $f' = 1$ everywhere.
• Expecting $\xi$ at the midpoint $\dfrac{\alpha + \beta}{2}$: the theorem says a $\xi$ exists, not where it is.

In the exam

• AP: “Is there a time in $(a, b)$ when the velocity equals …?” Name the theorem, state that the function is continuous on $[a, b]$ and differentiable on $(a, b)$, and compute the average rate of change — often from a table.
• Its consequences are exam staples: if $f' = 0$ on an interval then $f$ is constant, and if $f' > 0$ then $f$ is increasing — both are proved with this theorem.

Where it is taught

• AP Calculus AB/BC · Unit 5 (5.1 Using the Mean Value Theorem)
• Greek Γ΄ Λυκείου · §2.5–2.6