Show that the equation x - 17x +c = 0 has at most one solution in the interval [-2, 2]. Let f(x) = x - 17x + c for x in [-2, 2]. If f has two real solutions a and b in [-2, 2], with a < b, then --Select-- v. Since the polynomial f is continuous on [a, b] and differentiable on (a, b), --Select--- ) implies that there is a number r in (a, b) such that f'(r) = 0. Now f'(r) = . Since r is in (a, b), which is contained in [-2, 2], we have Irl < 2, so r² < 4. It follows that 3r2 - 17 ?v 3.4 – 17 = -5 ?v 0. This contradicts f'(r) = 0, so the given equation cannot have two real solutions in [-2, 2]. Hence, it has at most one real solution in [-2, 2].

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 3E
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Show that the equation x3 - 17x + c = 0 has at most one solution in the interval [-2, 2].
Let f(x) = x - 17x + c for x in [-2, 2]. If f has two real solutions a and b in [-2, 2], with a < b, then -Select---
polynomial f is continuous on [a, b] and differentiable on (a, b), -Select--
v. Since the
) implies that there is a number r in (a, b)
such that f'(r) = 0.
Now f'(r) =
. Since r is in (a, b), which is contained in [-2, 2], we have Irl < 2, so r? < 4. It follows that
3r2 - 17 ?v 3 ·4 - 17 = -5 ?v 0. This contradicts f'(r) = 0, so the given equation cannot have two real solutions in [-2, 2].
Hence, it has at most one real solution in [-2, 2].
Transcribed Image Text:Show that the equation x3 - 17x + c = 0 has at most one solution in the interval [-2, 2]. Let f(x) = x - 17x + c for x in [-2, 2]. If f has two real solutions a and b in [-2, 2], with a < b, then -Select--- polynomial f is continuous on [a, b] and differentiable on (a, b), -Select-- v. Since the ) implies that there is a number r in (a, b) such that f'(r) = 0. Now f'(r) = . Since r is in (a, b), which is contained in [-2, 2], we have Irl < 2, so r? < 4. It follows that 3r2 - 17 ?v 3 ·4 - 17 = -5 ?v 0. This contradicts f'(r) = 0, so the given equation cannot have two real solutions in [-2, 2]. Hence, it has at most one real solution in [-2, 2].
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