A continuous function on an interval can be drawn without lifting your pencil. That simple picture has a powerful consequence: such a function cannot skip any height between its endpoint values.
Idea clave
Intermediate Value Theorem (IVT). If $f$ is continuous on the closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there is at least one $c$ in $(a, b)$ with $f(c) = N$. In particular, if $f(a)$ and $f(b)$ have opposite signs, $f$ has a zero between them.
Por qué funciona
A full proof needs a property of the real numbers called completeness (the number line has no gaps). The idea: imagine the set of $x$ in $[a, b]$ where $f(x) < N$. Let $c$ be the edge of that set. Continuity forbids $f(c) < N$ (then slightly larger $x$ would also be below $N$) and forbids $f(c) > N$ (then slightly smaller $x$ would also be above $N$). What is left is $f(c) = N$.
Ejemplo resuelto: Locate a root
Show that $x^3 + x - 3 = 0$ has a solution in $[1, 2]$, then narrow it to an interval of length $0.1$.
- $f(x) = x^3 + x - 3$ is a polynomial, so it is continuous on $[1, 2]$.
- $f(1) = -1 < 0$ and $f(2) = 7 > 0$, so by the IVT there is a root in $(1, 2)$.
- Try more points: $f(1.2) = 1.728 + 1.2 - 3 = -0.072$ and $f(1.3) = 2.197 + 1.3 - 3 = 0.497$.
- The sign changes between 1.2 and 1.3, so a root lies in $(1.2, 1.3)$.
Respuesta: A root lies in $(1.2,\ 1.3)$. Repeating this halving idea pins it down as closely as we like.
Ejemplo resuelto: When the theorem does not apply
$f(x) = \frac{1}{x}$ has $f(-1) = -1$ and $f(1) = 1$. Does it have a zero in $(-1, 1)$?
- The signs differ, but $f$ is not continuous on $[-1, 1]$: it is undefined at $x = 0$.
- So the IVT says nothing. In fact $\frac{1}{x}$ is never 0: the graph jumps from $-\infty$ to $\infty$ across the asymptote.
Respuesta: No. The theorem's hypothesis fails.
Error común
Applying the IVT to a function that is not continuous on the whole closed interval, such as $\frac{1}{x}$ or a piecewise function with a jump. Before you conclude anything, check continuity on $[a, b]$ first. Also note what the theorem does not say: it promises at least one $c$, not exactly one, and it does not tell you where $c$ is.