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]. %3D (a) First, we show that f has a root in the interval (-4, –1). Since f is a choose 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]. f(-4) = and f(-1) = (-4, –1) by the choose (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 x = a and x = b in the interval [-4, –1] with a < b. Then f(a) = f(b) = . Since f is choose on the interval [-4, –1] and choose point c in interval (a, b) so that f' (c) = 0. However, the only solution to f' (x) = 0 is x = (a, b), since (a, b) C [-4, –1]. Thus, f cannot have more than one root in [-4, –1]. + on the interval (-4, –1), by choose + there would exist a which is not in the interval

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.2: Transformations Of Quadratic Functions
Problem 49PFA
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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].
%3D
(a) First, we show that f has a root in the interval (-4, –1). Since f is a choose
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].
f(-4) =
and f(-1) =
(-4, –1) by the choose
(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 x = a and x = b in the interval [-4, –1] with a < b. Then f(a) = f(b) =
. Since f is choose
on the interval [-4, –1] and choose
point c in interval (a, b) so that f' (c) = 0. However, the only solution to f' (x) = 0 is x =
(a, b), since (a, b) C [-4, –1]. Thus, f cannot have more than one root in [-4, –1].
+ on the interval (-4, –1), by choose
+ there would exist a
which is not in the interval
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]. %3D (a) First, we show that f has a root in the interval (-4, –1). Since f is a choose 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]. f(-4) = and f(-1) = (-4, –1) by the choose (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 x = a and x = b in the interval [-4, –1] with a < b. Then f(a) = f(b) = . Since f is choose on the interval [-4, –1] and choose point c in interval (a, b) so that f' (c) = 0. However, the only solution to f' (x) = 0 is x = (a, b), since (a, b) C [-4, –1]. Thus, f cannot have more than one root in [-4, –1]. + on the interval (-4, –1), by choose + there would exist a which is not in the interval
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