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Existence Theorem — Definition, Formula & Examples

An existence theorem is a theorem that guarantees something exists, such as a root, a maximum, or a point with a certain slope, as long as certain conditions are met. It tells you the object is there. It usually does not tell you where it is or how to find it.

An existence theorem is a statement of the form: if hypotheses HH hold, then there exists at least one object xx such that property P(x)P(x) is true. Symbolically, H⇒∃ x:P(x)H \Rightarrow \exists\, x : P(x). If the proof does not produce xx explicitly, it is called non-constructive. The conclusion guarantees existence, not uniqueness, unless the theorem says otherwise.

Worked Example

Problem: Show that the equation x3+x−1=0x^3 + x - 1 = 0 has at least one solution between x=0x = 0 and x=1x = 1.
Step 1: Check the hypothesis of the Intermediate Value Theorem. Let f(x)=x3+x−1f(x) = x^3 + x - 1. Every polynomial is continuous everywhere, so ff is continuous on the closed interval [0,1][0, 1].
f(x)=x3+x−1f(x) = x^3 + x - 1
Step 2: Evaluate ff at both endpoints.
f(0)=03+0−1=−1,f(1)=13+1−1=1f(0) = 0^3 + 0 - 1 = -1, \qquad f(1) = 1^3 + 1 - 1 = 1
Step 3: The target value 00 lies between f(0)=−1f(0) = -1 and f(1)=1f(1) = 1. By the IVT, some cc in (0,1)(0, 1) must satisfy f(c)=0f(c) = 0.
f(0)<0<f(1)  ⟹  ∃ c∈(0,1):f(c)=0f(0) < 0 < f(1) \implies \exists\, c \in (0,1) : f(c) = 0
Answer: The equation has at least one solution cc with 0<c<10 < c < 1. The theorem proves the solution exists but does not give its value. To find it, you need a separate method such as bisection or Newton's method.

Why It Matters

On the AP Calculus exam, free-response questions often ask you to justify that a value must exist using the IVT, EVT, or Mean Value Theorem. Full credit requires you to state that the hypotheses, usually continuity on a closed interval, are satisfied. Existence theorems also justify root-finding algorithms: bisection works only because the IVT guarantees a root inside each shrinking interval.

Common Mistakes

Mistake: Applying an existence theorem without checking its hypotheses. For example, you might note that f(x)=1xf(x) = \frac{1}{x} has f(−1)=−1f(-1) = -1 and f(1)=1f(1) = 1, then conclude that f(c)=0f(c) = 0 for some cc in (−1,1)(-1, 1).
Correction: The function 1x\frac{1}{x} is not continuous at x=0x = 0, so the IVT does not apply, and in fact 1x\frac{1}{x} is never 00. Always verify every condition, such as continuity or a closed interval, before using the conclusion.

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