Proving the Intermediate Value Theorem (IVT) for Calculus: Explained with Examples and Step-by-Step Proof

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

This statement is a special case of the Intermediate Value Theorem (IVT) in calculus

This statement is a special case of the Intermediate Value Theorem (IVT) in calculus. The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b] and takes on two values, f(a) and f(b), then it must also take on every value in between (including k).

To prove this, let’s assume that f(a) < k < f(b) (the same proof can be applied to the case f(a) > k > f(b)). Since f is a continuous function on the closed interval [a, b], it means that f(x) is defined and continuous for every x in the interval [a, b].

Using the IVT, we consider the function g(x) = f(x) – k. We can see that g(a) = f(a) – k < 0 and g(b) = f(b) - k > 0.

Now, since g(x) is continuous on [a, b] (as it is the difference of two continuous functions, f(x) and k), we can apply the Intermediate Value Theorem to g(x). The IVT guarantees that if a function is continuous on a closed interval and takes on two values, one positive and one negative, then it must also take on every value in between.

Therefore, there exists at least one number c in the interval [a, b] such that g(c) = 0, which means f(c) – k = 0. Rearranging this equation, we find f(c) = k.

In conclusion, if f is a continuous function on the closed interval [a, b] and k is a number between f(a) and f(b), then there exists at least one number c in the interval [a, b] such that f(c) = k.

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