Lesson 46 — Intermediate Value Theorem (IVT) and Applications

IVT and variants

Special case ($k = 0$): $f(a) \cdot f(b) < 0 \Rightarrow \exists c \in (a, b) : f(c) = 0$.

Application 1: Bisection (algorithm)

If $f(a) f(b) < 0$:

1. $m = (a+b)/2$.
2. If $f(m) = 0$, stop.
3. If $f(a) f(m) < 0$, repeat in $[a, m]$.
4. Otherwise, repeat in $[m, b]$.

Convergence: error halves per iteration, $|c - m_n| \leq (b-a)/2^n$.

Application 2: Fixed point

If $g \in C([a, b])$ with $g([a, b]) \subseteq [a, b]$, then $\exists c$ with $g(c) = c$. Apply IVT to $f(x) = g(x) - x$.

Application 3: Existence of equation solutions

$\cos x = x$, $e^x = 2x$, $\tan x = x$ have solutions by IVT.

Proof via completeness

Let $S = \{x \in [a, b] : f(x) < k\}$ (assuming $f(a) < k < f(b)$). $S \neq \emptyset$ ($a \in S$) and bounded above. Take $c = \sup S$. By continuity and supremum properties, $f(c) = k$.

Why continuity is needed

Counterexample: $f(x) = \mathrm{sgn}(x)$ on $[-1, 1]$. $f(-1) = -1$, $f(1) = 1$, but $f$ never takes the value $0.5$. Fails because $f$ is not continuous at 0.

IVT vs MVT

IVT: existence of an intermediate value ($f$ continuous). MVT (Mean Value Theorem, Lesson 56): existence of an intermediate derivative ($f$ differentiable).

Don't confuse them. MVT requires differentiability; IVT only continuity.