You Can't Climb Without Crossing Every Height

A continuous function on [a,b][a,b] must step on every value between f(a)f(a) and f(b)f(b). Move the target level cc and find the crossing ξ\xi.

f(a)<c<f(b)⇒∃ ξ∈(a,b):f(ξ)=cf(a)<c<f(b)\Rightarrow\exists\,\xi\in(a,b):f(\xi)=c

Examples & Exercises

Use the Intermediate Value Theorem to show that x3+x−1=0x^3 + x - 1 = 0 has a real root in (0,1)(0,1).

f(0)=−1<0<1=f(1) ⟹ ∃ ξ∈(0,1): f(ξ)=0f(0) = -1 < 0 < 1 = f(1) \ \Longrightarrow\ \exists\, \xi \in (0,1):\ f(\xi) = 0
c=0c = 0 — the point ξ≈0.6823\xi \approx 0.6823 where the curve steps on height 00.
Show proof
  1. f(x)=x3+x−1f(x) = x^3 + x - 1 is a polynomial, hence continuous on [0,1][0,1].
  2. f(0)=−1<0f(0) = -1 < 0 and f(1)=1>0f(1) = 1 > 0.
  3. By the Intermediate Value Theorem there is ξ∈(0,1)\xi \in (0,1) with f(ξ)=0f(\xi) = 0 (strictly inside, since f(0),f(1)≠0f(0), f(1) \neq 0). Numerically ξ≈0.6823278038\xi \approx 0.6823278038.

Show that sin⁡x=12\sin x = \frac12 has a solution in (0,π2)(0, \frac{\pi}{2}).

sin⁡0=0<12<1=sin⁡π2 ⟹ ∃ ξ: sin⁡ξ=12\sin 0 = 0 < \tfrac12 < 1 = \sin \tfrac{\pi}{2} \ \Longrightarrow\ \exists\, \xi:\ \sin \xi = \tfrac12
c=0.5c = 0.5 — the crossing is at ξ=π/6≈0.5236\xi = \pi/6 \approx 0.5236.
Show proof
  1. sin⁡x\sin x is continuous on [0,π2][0, \frac{\pi}{2}].
  2. sin⁡0=0<12\sin 0 = 0 < \frac12 and sin⁡π2=1>12\sin \frac{\pi}{2} = 1 > \frac12.
  3. By the Intermediate Value Theorem there is ξ∈(0,π2)\xi \in (0, \frac{\pi}{2}) with sin⁡ξ=12\sin \xi = \frac12. For reference, sin⁡π6=12\sin \frac{\pi}{6} = \frac12, so in fact ξ=π6\xi = \frac{\pi}{6} — the theorem guarantees existence; the exact value is a separate computation.

Show directly that f(x)={0x<121x≥12f(x) = \begin{cases} 0 & x < \frac12 \\ 1 & x \geq \frac12 \end{cases} on [0,1][0,1] never takes the value c=12c = \frac12, and name the IVT hypothesis that fails.

f([0,1])={0,1}∌12f([0,1]) = \{0, 1\} \not\ni \tfrac12
The continuous case (c=0c = 0) — drop the hypothesis and the crossing may vanish.
Show proof
  1. For x<12x < \frac12, f(x)=0≠12f(x) = 0 \neq \frac12; for x≥12x \geq \frac12, f(x)=1≠12f(x) = 1 \neq \frac12. So no xx satisfies f(x)=12f(x) = \frac12, even though f(0)=0<12<1=f(1)f(0) = 0 < \frac12 < 1 = f(1).
  2. The broken hypothesis is continuity at x=12x = \frac12: the left-hand limit is 00 while f(12)=1f(\frac12) = 1. This counterexample shows the IVT conclusion need not hold without its hypothesis.

3. Continuity