Suppose f(x) = x° + 3x + 1. In this problem, we will show that f has exactly one root (or zero) in the interval [-2, -1]. (a) First, we show that f has a root in the interval (-2, –1). Since f is a continuous function on the interval [-2, -1] and f(-2) =| 59 and f(-1) =|-1 the graph of y = f(x) must cross the -axis at some point in the interval by the intermediate value theorem . Thus, f has at least one root in the interval [-2, -1]. (b) Second, we show that f cannot have more than one root in the interval [-2, –1] by a thought b in the interval [-2, –1] with experiment. Suppose that there were two roots x = a and x = a < b. Then f(a) = f(b) =| Since f is continuous differentiable on the interval (-2, –1), by Rolle's theorem + on the interval [-2, –1] and there would exist a point c in interval (a, b) so that f'(c) = 0. However, the only solution to f'(x) = 0 is x = %3D which is not in the interval (a, b), since (a, b) C[-2, –1]. Thus, f cannot have more than one root in [-2, –1].

College Algebra (MindTap Course List)
12th Edition
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Author:R. David Gustafson, Jeff Hughes
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Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
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Precalculus
Suppose f(x) = x° + 3x + 1. In this problem, we will show that f has exactly one root (or zero) in
the interval [-2, –1].
(a) First, we show that f has a root in the interval (-2, -1). Since f is a continuous
function on the interval [-2, -1] and f(-2) = | 59 and f(-1) =|-1
the graph of y = f(x)
must cross the -axis at some point in the interval
by the
intermediate value theorem . Thus, f has at least one root in the interval [-2, –1].
(b) Second, we show that f cannot have more than one root in the interval [-2, –1] by a thought
b in the interval [-2, –1] with
experiment. Suppose that there were two roots x = a and x =
a < b. Then f(a) = f(b) =| |Since f is continuous
differentiable on the interval (-2, –1), by Rolle's theorem
exist a point c in interval (a, b) so that f'(c) = 0. However, the only solution to f'(x) = 0 is x =
which is not in the interval (a, b), since (a, b) C [-2, – 1]. Thus, f cannot have more than one
+ on the interval [-2, –1] and
there would
%3D
root in [-2, –1].
Transcribed Image Text:Suppose f(x) = x° + 3x + 1. In this problem, we will show that f has exactly one root (or zero) in the interval [-2, –1]. (a) First, we show that f has a root in the interval (-2, -1). Since f is a continuous function on the interval [-2, -1] and f(-2) = | 59 and f(-1) =|-1 the graph of y = f(x) must cross the -axis at some point in the interval by the intermediate value theorem . Thus, f has at least one root in the interval [-2, –1]. (b) Second, we show that f cannot have more than one root in the interval [-2, –1] by a thought b in the interval [-2, –1] with experiment. Suppose that there were two roots x = a and x = a < b. Then f(a) = f(b) =| |Since f is continuous differentiable on the interval (-2, –1), by Rolle's theorem exist a point c in interval (a, b) so that f'(c) = 0. However, the only solution to f'(x) = 0 is x = which is not in the interval (a, b), since (a, b) C [-2, – 1]. Thus, f cannot have more than one + on the interval [-2, –1] and there would %3D root in [-2, –1].
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