Show that the equation x^4 + 4x + c = 0 has at most two real roots.

I believe we're supposed to prove this by proof of contradiction using Rolle's Theorem, but I'm not quite sure how to do this problem.

To show that the equation x^4 + 4x + c = 0 has at most two real roots, we can indeed use a proof by contradiction along with Rolle's Theorem.

First, let's assume that the equation has more than two real roots. That means there are at least three real roots, which we'll call r1, r2, and r3 (possibly with r1 = r2 or r2 = r3).

Rolle's Theorem states that if a function is continuous in a closed interval [a, b], differentiable in the open interval (a, b), and if the function takes the same value at both endpoints, then there exists at least one point within the interval where the derivative of the function is equal to zero.

Now, let's define a function f(x) = x^4 + 4x + c, which represents the left-hand side of the equation.

Since the equation has at least three real roots, we know that f(r1) = f(r2) = f(r3) = 0.

Using Rolle's Theorem, we can conclude that between any two distinct roots, there must exist a point where the derivative of f(x) is equal to zero.

Let's consider the interval between r1 and r2. By Rolle's Theorem, there exists a point in this interval, let's call it x1, such that f'(x1) = 0.

Similarly, considering the interval between r2 and r3, there exists another point x2 in that interval where f'(x2) = 0.

Now, let's find the derivative of f(x) = x^4 + 4x + c. Taking the derivative, we get f'(x) = 4x^3 + 4.

Setting f'(x1) = 0, we have:
4x1^3 + 4 = 0
4x1^3 = -4
x1^3 = -1

This means x1 is a real number such that its cube is equal to -1. However, no real number cubed can be equal to -1. This contradicts our assumption that more than two real roots exist.

Hence, we can conclude that the equation x^4 + 4x + c = 0 has at most two real roots.