Darboux Theorem

The Darboux theorem, also known as the intermediate value property, states that if a function f(x) is differentiable on an interval [a, b], then it satisfies the intermediate value property. In other words, if f(a) and f(b) have opposite signs, then there exists at least one point c in the interval (a, b) such that f(c) = 0.

Proof:

To prove the Darboux theorem, we can consider two cases:

Case 1: f(a) < 0="" and="" f(b)="" /> 0

In this case, we can define a new function g(x) = f(x) - x. Since f(x) is differentiable on [a, b], g(x) is also differentiable on the same interval.

Claim 1: g(a) < 0="" and="" g(b)="" /> 0

Proof of Claim 1:
Since f(a) < 0="" and="" f(b)="" /> 0, we have g(a) = f(a) - a < 0="" and="" g(b)="f(b)" -="" b="" /> 0.

Claim 2: There exists a point c in (a, b) such that g'(c) = 0

Proof of Claim 2:
Since g(x) is differentiable on [a, b], it satisfies the conditions of the Rolle's theorem. Therefore, there must exist a point c in (a, b) such that g'(c) = 0.

Now, let's analyze g'(c):
g'(c) = f'(c) - 1

Since g'(c) = 0, we have f'(c) = 1.

Claim 3: There exists a point d in (a, b) such that f(d) = 0

Proof of Claim 3:
Using the mean value theorem, we know that if f'(c) = 1, then there exists a point d in (a, b) such that f(d) = f(c) + f'(c)(d - c) = 0.

Therefore, we have proved that if f(a) < 0="" and="" f(b)="" /> 0, then there exists at least one point d in (a, b) such that f(d) = 0.

Case 2: f(a) > 0 and f(b) < />

This case can be proven similarly to Case 1 by considering the function -f(x) instead of f(x). By applying the same steps, we can show that there exists at least one point c in (a, b) such that f(c) = 0.

Hence, we have proved the Darboux theorem, which states that if a function f(x) is differentiable on an interval [a, b] and f(a) and f(b) have opposite signs, then there exists at least one point c in the interval (a, b) such that f(c) = 0.