Strategy: We need to do two things here. First, we show that the equation has a solution. Then we will need to show that there cannot be two solutions. We will use the Intermediate Value Theorem for the existence of a solution, and Rolle’s Theorem (or the MVT) for the fact that there is only one. Our strategy will include the idea that for the function f(x) = x3 + ex, this problem is equivalent to saying that the graph of f(x) has only one x-intercept.
Solution: First note that the function f (x) = x3 + ex is differentiable on all of R, since it is the sum of two differentiable functions, a polynomial and the exponential function. Hence it is continuous. Grabbingthetwoinputvaluesx=−1andx=0,weseethatf(−1)=(−1)3+e−1 =−1+1e <0, and f (0) = (0)3 + e0 = 1 > 0. Hence by the Intermediate Value Theorem, we know that there must be a number c ∈ [−1, 0], where f (c) = 0. But this is our solution to the original equation. Hence a solution exists.
So call c our solution to the original equation and suppose there exists another one; Call this one d. Then either d > c or d < c. It really won’t matter for our discussion, so assume d > c. Then f(x) is a continuous function on [c,d] and differentiable on (c,d) (it is differentiable everywhere, actually). By Rolle’s Theorem, there then must exist a point r ∈ (c,d), where f′(r) = 0. However, we have
f′(x) = 3x2 + ex > 0
for all x ∈ R. Thus, like our example in class, there cannot be a point r ∈ (c,d), where f′(r) = 0. So our assumption that there is a second solution to the original equation is incorrect. Thus there is only one real solution to x3 + ex = 0.