15. [-/8 Points] DETAILS MY NOTES Show that the equation has exactly one real root. x5+ ex = 0 SESSCALCET2 4.2.018. Let f(x)=x5+ ex. Then f(-1)= ?00 and f(0) = ?0. Since f is the sum of a polynomial and the natural exponential function, f is continuous and differentiable for all x. By the Intermediate Value Theorem, there is a number c in (-1, 0) such that f(c) 0. Thus, the given equation has at least one real root. If the equation has distinct real roots a and b with a < b, then f(a) = f(b)=0. Since f is continuous on [a, b] and differentiable on (a, b), Rolle's Theorem implies that there is a number r in (a, b) such that ?0. This contradiction shows that the given equation can't have two distinct real roots, so it has exactly one root. f(r) 0. But f'(r) = Need Help? Read It