Extreme value theorem

What you are looking at

The working gives $M$ and $m$, where they occur, and the range $f([\alpha, \beta]) = [m, M]$, drawn as a green bar on the $y$-axis. The maximum and minimum can sit inside the interval, at a turning point, or at an endpoint — drag $\alpha$ and $\beta$ and watch them move from one to the other.

Choose Open and the endpoints are removed. If the largest value was at an endpoint it is no longer attained: the values get as close to it as you like but never reach it, so there is a supremum but no maximum, and the range becomes half-open. Continuity alone is not enough; the interval has to be closed and bounded.

Try this

1. 1On $f(x) = 2\sin x + \cos 2x$ over $[-2, 3]$, the maximum $1.5$ is attained twice, at $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$: the maximum value is unique, the points where it happens need not be.
2. 2Move the interval to $[-1, 0]$, where $f$ is increasing: the minimum is at $\alpha$ and the maximum at $\beta$.
3. 3Now choose Open $(\alpha, \beta)$: both extremes were at the endpoints, so neither is attained any more.
4. 4Switch to $x^3 - 3x + 1$ on $[-1.5, 2]$: the maximum $3$ is reached both at the turning point $x = -1$ and at the endpoint $x = 2$.

Mistakes students make

• Checking only the turning points. A maximum or minimum can be at an endpoint — always compare the critical values with $f(\alpha)$ and $f(\beta)$.
• Using it on an open interval or across a discontinuity: $f(x) = x$ on $(0, 1)$ has no maximum, and $\dfrac{1}{x}$ on $(0, 1]$ has none either.
• Confusing the maximum value with where it happens: the value $M$ is unique, the points $x$ with $f(x) = M$ need not be.

In the exam

• AP’s candidates test: find the critical points, evaluate $f$ there and at both endpoints, and justify the absolute maximum by comparing the values. The theorem is why a closed interval always has an answer.
• Optimisation on a closed domain, at any level, needs the endpoint check for the same reason.

Where it is taught

• AP Calculus AB/BC · Unit 5 (5.2 Extreme Value Theorem, 5.5 candidates test)
• IB Maths AA SL/HL · 5.8 (local and global extrema, optimisation)
• Greek Γ΄ Λυκείου · §1.8