The Extreme Value Theorem

A continuous function on a closed interval always attains its max and min. Verify it for f(x)=x3−3xf(x) = x^3 - 3x on [−2,2][-2,2], and see from the open-interval and discontinuous counterexamples that both hypotheses are necessary.

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]: 0≤x1<x2≤10 \leq x_1 < x_2 \leq 1 implies 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, x=u+12∈[0,1)x = \frac{u+1}{2} \in [0,1) gives f(x)=u+12>uf(x) = \frac{u+1}{2} > u, so uu is not an upper bound. Hence 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.

3. Continuity