The Extreme Value Theorem

This scene checks the extreme value theorem. On a closed interval the maximum is attained; open intervals and breaks show why it can fail.

Two conditions are needed. The interval must be closed, and the graph must be unbroken. Drop either one, and the highest point may only be approached, never reached. That is why both hypotheses matter.

For example, picture a 🧑 hiker walking a trail to the summit. If the trail runs all the way, some step lands on the highest point. If the trail breaks just before it, the hiker gets close but never arrives.

f continuous on [a,b]⇒∃ xmax⁡,xmin⁡f \text{ continuous on } [a,b] \Rightarrow \exists\, x_{\max}, x_{\min}
f(x)=x3−3x,f(−1)=f(2)=2, f(−2)=f(1)=−2f(x) = x^3 - 3x,\quad f(-1) = f(2) = 2,\ f(-2) = f(1) = -2

Part of these learning paths

Examples & Exercises

Find the maximum and minimum of x2x^2 on [0,1][0,1], and say where each is attained.

max⁡[0,1]x2=1 at x=1,min⁡[0,1]x2=0 at x=0\max_{[0,1]} x^2 = 1\ \text{at } x = 1, \qquad \min_{[0,1]} x^2 = 0\ \text{at } x = 0
Closed mode — as the cubic attains both extremes, so does x2x^2 on [0,1][0,1].
Show proof

  1. x2x^2 is increasing on [0,1][0,1]: if 0≤x1<x2≤10 \leq x_1 < x_2 \leq 1, factor x22−x12=(x2−x1)(x1+x2)x_2^2 - x_1^2 = (x_2 - x_1)(x_1 + x_2); here x2−x1>0x_2 - x_1 > 0 and x1+x2>0x_1 + x_2 > 0 (the case x1=x2=0x_1 = x_2 = 0 is excluded by x1<x2x_1 < x_2, and if x1=0<x2x_1 = 0 < x_2 then x1+x2=x2>0x_1 + x_2 = x_2 > 0). Hence the product is positive, so x12<x22x_1^2 < x_2^2.

  2. Hence 0=02≤x2≤12=10 = 0^2 \leq x^2 \leq 1^2 = 1 for all x∈[0,1]x \in [0,1]. The max 11 is attained at x=1x = 1; the min 00 at x=0x = 0.

  3. Consistent with the Extreme Value Theorem (cited): [0,1][0,1] is closed and x2x^2 is continuous, so both extremes must exist.

Prove f(x)=1/xf(x) = 1/x has no maximum on (0,1](0,1]; name the hypothesis that fails.

f ⁣(1M+1)=M+1>M∀Mf\!\left(\frac{1}{M+1}\right) = M + 1 > M \qquad \forall M
Open-interval mode — the curve escapes upward without bound.
Show proof

  1. Suppose MM were a maximum.

    If M<0M < 0, then f(1)=1>Mf(1) = 1 > M, already a contradiction.

    If M≥0M \geq 0, put x=1M+1x = \frac{1}{M+1}: then 0<x≤10 < x \leq 1, so x∈(0,1]x \in (0,1], and f(x)=M+1>Mf(x) = M + 1 > M — a contradiction.

  2. Hence no MM can be an upper bound: ff is unbounded above, so no maximum exists.

  3. The broken hypothesis is the closedness of the domain: (0,1](0,1] is not a closed interval. As x→0+x \to 0^+ there is an open "escape hole" through which ff diverges.

For f(x)=xf(x) = x on [0,1)[0,1) with f(1)=0f(1) = 0, prove sup⁡f=1\sup f = 1 but 11 is never attained; name the hypothesis that fails.

sup⁡[0,1]f=1∉f([0,1])\sup_{[0,1]} f = 1 \notin f([0,1])
Discontinuous mode — the value 11 is approached but never touched.
Show proof

  1. For x∈[0,1)x \in [0,1), f(x)=x<1f(x) = x < 1; and f(1)=0f(1) = 0. So f(x)<1f(x) < 1 for every xx — no point takes the value 11.

  2. sup⁡=1\sup = 1: 11 is an upper bound (since f(x)<1f(x) < 1).

    For any real u<1u < 1, produce x∈[0,1)x \in [0,1) with f(x)>uf(x) > u.

    If u<0u < 0, take x=0x = 0: f(0)=0>uf(0) = 0 > u.

    If 0≤u<10 \le u < 1, take x=u+12x = \frac{u+1}{2}: then u≤u+12<1u \le \frac{u+1}{2} < 1, so x∈[0,1)x \in [0,1) and f(x)=u+12>uf(x) = \frac{u+1}{2} > u.

    (Note: for u<−1u < -1, u+12<0\frac{u+1}{2} < 0 falls outside [0,1)[0,1), which is why x=0x = 0 is needed there.)

    Hence no number below 11 is an upper bound, and the least upper bound is 11.

  3. The broken hypothesis is continuity at x=1x = 1: the left-hand limit is 11 but f(1)=0f(1) = 0. Even on a closed interval, dropping continuity can leave the maximum as an unattained supremum.

4. Continuity