Intermediate Value Theorem (IVT): the 3-step method

Statement of IVT

If $f$ is continuous on $[a, b]$ and $k$ is between $f(a)$ and $f(b)$, then there exists at least one $c \in [a, b]$ such that $f(c) = k$.

In plain words: a continuous function “takes every value between” $f(a)$ and $f(b)$.

Special case: $k = 0$

If $f$ is continuous on $[a, b]$ and $f(a) \times f(b) < 0$ (opposite signs), then $f$ vanishes at least once on $(a, b)$.

Corollary (bijection): uniqueness

If $f$ is continuous AND strictly monotonic on $[a, b]$, and $k$ lies between $f(a)$ and $f(b)$, then there exists a UNIQUE $c \in [a, b]$ such that $f(c) = k$.

3-step method

To show that $f(x) = k$ has a unique solution on $[a, b]$:

1. Continuity: state that $f$ is continuous (often because differentiable).
2. Strict monotonicity: study $f'$, say ”$f$ is strictly increasing on $[a, b]$”.
3. Bracketing: check $f(a) < k < f(b)$ (or the reverse). Conclude by the corollary.

Typical example

Show that $x^3 + x - 1 = 0$ has a unique solution in $[0, 1]$.

Let $f(x) = x^3 + x - 1$.

1. $f$ is a polynomial ⇒ continuous on $[0, 1]$.
2. $f'(x) = 3x^2 + 1 > 0$ ⇒ $f$ strictly increasing.
3. $f(0) = -1 < 0 < 1 = f(1)$.

Corollary of IVT ⇒ unique solution in $(0, 1)$.

Localizing the solution (bisection)

Once existence is proven, we can approximate the solution by cutting the interval in half and keeping the one where $f$ changes sign. This is the bisection method.

Common mistakes

1. Forgetting continuity (essential: without it, IVT fails).
2. Using plain IVT for uniqueness (it only gives existence).
3. Confusing $f$ monotonic and $f'$ monotonic.
4. Skipping the bracketing check $f(a) < k < f(b)$.

?À toi de jouer

IVT + strict monotonicity give:

A.Existence AND uniquenessB.Existence onlyC.Uniqueness onlyD.Nothing certain

Valider