
# 7.2: Proof of the Intermediate Value Theorem


Skills to Develop

• Proof of Intermediate value theorem

We now have all of the tools to prove the Intermediate Value Theorem (IVT).

Theorem $$\PageIndex{1}$$: Intermediate Value Theorem

Suppose $$f(x)$$ is continuous on $$[a,b]$$ and v is any real number between $$f(a)$$ and $$f(b)$$. Then there exists a real number $$c ∈ [a,b]$$ such that $$f(c) = v$$.

Sketch of Proof

We have two cases to consider: $$f(a) ≤ v ≤ f(b)$$ and $$f(a) ≥ v ≥ f(b)$$.

We will look at the case $$f(a) ≤ v ≤ f(b)$$. Let $$x_1 = a$$ and $$y_1 = b$$, so we have $$x_1 ≤ y_1$$ and $$f(x_1) ≤ v ≤ f(y_1)$$. Let $$m_1$$ be the midpoint of $$[x_1,y_1]$$ and notice that we have either $$f(m_1) ≤ v$$ or $$f(m_1) ≥ v$$. If $$f(m_1) ≤ v$$, then we relabel $$x_2 = m_1$$ and $$y_2 = y_1$$. If $$f(m_1) ≥ v$$, then we relabel $$x_2 = x_1$$ and $$y_2 = m_1$$. In either case, we end up with $$x_1 ≤ x_2 ≤ y_2 ≤ y_1,\; y_2 - x_2 = \frac{1}{2} (y_1 - x_1)$$, $$f(x_1) ≤ v ≤ f(y_1)$$, and $$f(x_2) ≤ v ≤ f(y_2)$$.

Now play the same game with the interval $$[x_2,y_2]$$. If we keep playing this game, we will generate two sequences ($$x_n$$) and ($$y_n$$), satisfying all of the conditions of the nested interval property. These sequences will also satisfy the following extra property: $$∀ n,\; f(x_n) ≤ v ≤ f(y_n)$$. By the NIP, there exists a $$c$$ such that $$x_n ≤ c ≤ y_n,\; ∀ n$$. This should be the $$c$$ that we seek though this is not obvious. Speciﬁcally, we need to show that $$f(c) = v$$. This should be where the continuity of $$f$$ at $$c$$ and the extra property on ($$x_n$$)and ($$y_n$$) come into play.

Exercise $$\PageIndex{1}$$

Turn the ideas of the previous paragraphs into a formal proof of the IVT for the case $$f(a) ≤ v ≤ f(b)$$.

Exercise $$\PageIndex{2}$$

We can modify the proof of the case $$f(a) ≤ v ≤ f(b)$$ into a proof of the IVT for the case $$f(a) ≥ v ≥ f(b)$$. However, there is a sneakier way to prove this case by applying the IVT to the function $$-f$$. Do this to prove the IVT for the case $$f(a) ≥ v ≥ f(b)$$.

Exercise $$\PageIndex{3}$$

Use the IVT to prove that any polynomial of odd degree must have a real root.

## Contributor

• Eugene Boman (Pennsylvania State University) and Robert Rogers (SUNY Fredonia)