Suppose f(x) = x° + 3x + 1. In this problem, we will show that f has exactly one root (or zero) in the interval [-4, –1]. [- a) First, we show that f has a root in the interval (-4, –1). Since f is a f(-4) = function on the interval [-4, -1] and the graph of y = f(x) must cross the x-axis at some point in the interval Thus, f has at least one root in the interval [-4, –1]. and f(-1) = (-4, –1) by the (b) Second, we show that f cannot have more than one root in the interval [-4, –1] by a thought experiment. Suppose that there were two roots = a and x = b in the interval [-4, –1] with a < b. Then f(a) = f(b) = on the interval (-4, –1), by .Since f is on the interval [-4, –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 = which is not in the interval (a, b), since (a, b) C [-4, –1]. Thus, f cannot have more than one root in [-4, –1].
Suppose f(x) = x° + 3x + 1. In this problem, we will show that f has exactly one root (or zero) in the interval [-4, –1]. [- a) First, we show that f has a root in the interval (-4, –1). Since f is a f(-4) = function on the interval [-4, -1] and the graph of y = f(x) must cross the x-axis at some point in the interval Thus, f has at least one root in the interval [-4, –1]. and f(-1) = (-4, –1) by the (b) Second, we show that f cannot have more than one root in the interval [-4, –1] by a thought experiment. Suppose that there were two roots = a and x = b in the interval [-4, –1] with a < b. Then f(a) = f(b) = on the interval (-4, –1), by .Since f is on the interval [-4, –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 = which is not in the interval (a, b), since (a, b) C [-4, –1]. Thus, f cannot have more than one root in [-4, –1].
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 54E
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![Suppose f(x) = x° + 3x + 1. In this problem, we will show that f has exactly one root (or zero) in the interval [-4, –1].
[-
a) First, we show that f has a root in the interval (-4, –1). Since f is a
f(-4) =
function on the interval [-4, -1] and
the graph of y = f(x) must cross the x-axis at some point in the interval
Thus, f has at least one root in the interval [-4, –1].
and f(-1) =
(-4, –1) by the
(b) Second, we show that f cannot have more than one root in the interval [-4, –1] by a thought experiment. Suppose that there were
two roots = a and x =
b in the interval [-4, –1] with a < b. Then f(a) = f(b) =
on the interval (-4, –1), by
.Since f is
on the interval [-4, –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 =
which is not in the interval
(a, b), since (a, b) C [-4, –1]. Thus, f cannot have more than one root in [-4, –1].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f8ca34a-b915-479b-b71c-2aa6f3c43f7e%2Fe428b959-d2d8-4561-8a54-f6aa704aeb61%2Fff93us.png&w=3840&q=75)
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 [-4, –1].
[-
a) First, we show that f has a root in the interval (-4, –1). Since f is a
f(-4) =
function on the interval [-4, -1] and
the graph of y = f(x) must cross the x-axis at some point in the interval
Thus, f has at least one root in the interval [-4, –1].
and f(-1) =
(-4, –1) by the
(b) Second, we show that f cannot have more than one root in the interval [-4, –1] by a thought experiment. Suppose that there were
two roots = a and x =
b in the interval [-4, –1] with a < b. Then f(a) = f(b) =
on the interval (-4, –1), by
.Since f is
on the interval [-4, –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 =
which is not in the interval
(a, b), since (a, b) C [-4, –1]. Thus, f cannot have more than one root in [-4, –1].
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