Application of Rolle's Theorem

Have you ever looked at a polynomial and wondered if it has a root in a specific interval, without actually having to solve it?

Calculus gives us an incredibly elegant workaround for this exact scenario: Rolle's Theorem. Let's walk through a classic problem that demonstrates how we can find hidden roots by stepping up one level of abstraction.

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The Problem

Suppose we have four real numbers \(a_0, a_1, a_2,\) and \(a_3\) that satisfy the following condition:

$$a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \frac{a_3}{4} = 0$$

The Goal: Prove that the polynomial \(f(x) = a_0 + a_1x + a_2x^2 + a_3x^3\) has at least one root in the interval \((0, 1)\).

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The Intuition: Where do we start?

When asked to prove that a function \(f(x)\) equals zero at some point, a calculus student's first instinct should be to look for an antiderivative.

Why? Because Rolle's Theorem tells us that if a continuous, differentiable function starts and ends at the exact same height on a graph (for instance, if it equals \(0\) at both ends of an interval), it must flatten out somewhere in between. At that flat point, its derivative is exactly zero.

So, if we can find a parent function \(g(x)\) such that \(g'(x) = f(x)\), and we can show that \(g(x)\) is tied down at \(x=0\) and \(x=1\), the theorem will do the heavy lifting for us.

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The Proof

Step 1: Construct the parent function, \(g(x)\).
To find our function \(g(x)\), we simply integrate \(f(x)\) with respect to \(x\):

$$g(x) = \int (a_0 + a_1x + a_2x^2 + a_3x^3) \,dx$$
$$g(x) = a_0x + \frac{a_1}{2}x^2 + \frac{a_2}{3}x^3 + \frac{a_3}{4}x^4$$

(Note: We can ignore the constant of integration \(+C\) here by simply setting it to \(0\), as we just need one valid parent function).

Step 2: Evaluate the endpoints of our interval \([0, 1]\).
Let's plug our bounds into \(g(x)\) to see how the function behaves at the edges.

At \(x = 0\):

$$g(0) = a_0(0) + \frac{a_1}{2}(0)^2 + \frac{a_2}{3}(0)^3 + \frac{a_3}{4}(0)^4 = 0$$

At \(x = 1\):

$$g(1) = a_0(1) + \frac{a_1}{2}(1)^2 + \frac{a_2}{3}(1)^3 + \frac{a_3}{4}(1)^4$$
$$g(1) = a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \frac{a_3}{4}$$

Notice something familiar? The problem explicitly states that \(a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \frac{a_3}{4} = 0\). Therefore:

$$g(1) = 0$$

Step 3: Apply Rolle's Theorem.
We now have a function \(g(x)\) where:

1. \(g(x)\) is a polynomial, meaning it is perfectly continuous on the closed interval \([0,1]\) and differentiable on the open interval \((0,1)\).
2. \(g(0) = g(1) = 0\).

Because both endpoints are at the same \(y\)-value, the graph of \(g(x)\) must turn around at least once between \(x=0\) and \(x=1\).

By Rolle's Theorem, there must exist at least one point \(c\) in the interval \((0, 1)\) where the derivative is zero:

$$g'(c) = 0$$

Since \(g'(x)\) is just our original polynomial \(f(x)\), this means:

$$f(c) = a_0 + a_1c + a_2c^2 + a_3c^3 = 0$$

Conclusion:

We have successfully proven that \(f(x)\) has at least one root exactly where we wanted it: inside the interval \((0, 1)\).