Continuity and IVT: the tool that proves an equation has a solution

The theorem

Intermediate Value Theorem (IVT) — If $f$ is continuous on $[a, b]$, then for any $k$ between $f(a)$ and $f(b)$, there exists at least one $c \in [a, b]$ with $f(c) = k$.

The “has a zero” corollary

If $f$ is continuous on $[a, b]$ and $f(a)$, $f(b)$ have opposite signs, then there is at least one $c \in [a, b]$ with $f(c) = 0$.

If additionally $f$ is strictly monotonic, that $c$ is unique.

Method

To prove $f(x) = 0$ has a solution on $[a, b]$:

1. Check continuity (polynomials, sin, cos, exp, ln are continuous on their domain).
2. Compute $f(a)$ and $f(b)$.
3. Verify opposite signs (product < 0).
4. Conclude via IVT.
5. Optional: uniqueness via strict monotonicity ($f'$ sign).

Example

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

• $f(x) = x^3 + x - 1$ is polynomial → continuous on $[0, 1]$.
• $f(0) = -1 < 0$, $f(1) = 1 > 0$: opposite signs.
• IVT ⇒ some $c \in [0, 1]$ with $f(c) = 0$.
• $f'(x) = 3x^2 + 1 > 0$: $f$ is strictly increasing.
• ⇒ unique solution.

Common mistakes

1. Forgetting continuity: IVT needs continuity.
2. Existence vs uniqueness: IVT gives at least one $c$, not exactly one.
3. Requiring strictly opposite signs: $f(a) \cdot f(b) < 0$.
4. One interval isn’t enough: sometimes split $[a, b]$ into subintervals.