Show that the equation has exactly one real root. x5 + ex = 0 Let f(x) = x5 + eX. Then (-1) = 0 and (o) = 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 fr) = 0. But f'(r) = 0. This contradiction shows that the given equation can't have two distinct real roots, so it has exactly one root.
Show that the equation has exactly one real root. x5 + ex = 0 Let f(x) = x5 + eX. Then (-1) = 0 and (o) = 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 fr) = 0. But f'(r) = 0. This contradiction shows that the given equation can't have two distinct real roots, so it has exactly one root.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Show that the equation has exactly one real root.
x5 + ex = 0
Let f(x) = x5 + eX. Then (-1) =
0 and (o) =
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 fr) = 0. But
f'(r) =
0. This contradiction shows that the given equation can't
have two distinct real roots, so it has exactly one root.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3524539-73d7-4b1e-9607-006acd5ee568%2Fc895cf84-ce14-4226-8d09-e8046b1d39f1%2Felxxtmc.jpeg&w=3840&q=75)
Transcribed Image Text:Show that the equation has exactly one real root.
x5 + ex = 0
Let f(x) = x5 + eX. Then (-1) =
0 and (o) =
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 fr) = 0. But
f'(r) =
0. This contradiction shows that the given equation can't
have two distinct real roots, so it has exactly one root.
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