Suppose f(x) = =°+ 3z + 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 v function on the interval [-2, -1] and f(-2) = and f(-1) = the graph of y = f(x) must cross the r-axis at some point in the interval (-2, –1) by the (두)- intermediate value theorem v 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 experiment. Suppose that there were two roots z = a and z = b in the interval [-2, -1] with a < b. Then f(a) = f(b) =| | Sincef is continuous v on the interval -2,-1 and differentiable v on the interval (-2,-1), by Rolle's theorem v 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 = more than one root in -2, –1. which is not in the interval (a, b), since (a, b) C[-2,-1. Thus, f cannot have (Note: where the problem asks you to make a choice select the weakest choice that works in the given context. For example "continuous" is a weaker condition than "polynomial" because every polynomial is continuous but not vice-versa. Rolle's theorem is a weaker theorem than the mean value theorem because Rolle's theorem applies to fewer cases.)
Suppose f(x) = =°+ 3z + 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 v function on the interval [-2, -1] and f(-2) = and f(-1) = the graph of y = f(x) must cross the r-axis at some point in the interval (-2, –1) by the (두)- intermediate value theorem v 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 experiment. Suppose that there were two roots z = a and z = b in the interval [-2, -1] with a < b. Then f(a) = f(b) =| | Sincef is continuous v on the interval -2,-1 and differentiable v on the interval (-2,-1), by Rolle's theorem v 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 = more than one root in -2, –1. which is not in the interval (a, b), since (a, b) C[-2,-1. Thus, f cannot have (Note: where the problem asks you to make a choice select the weakest choice that works in the given context. For example "continuous" is a weaker condition than "polynomial" because every polynomial is continuous but not vice-versa. Rolle's theorem is a weaker theorem than the mean value theorem because Rolle's theorem applies to fewer cases.)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
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ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
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Chapter1: Functions
Section1.2: Functions Given By Tables
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![Suppose f(x) = =°+ 3z + 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
v function on the interval [-2, -1] and f(-2) =
and f(-1) =|
the graph of y = f(x) must cross the r-axis at some point in the interval (-2, –1) by the
intermediate value theorem v
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 experiment. Suppose that there were two
roots z = a and z = b in the interval [-2, -1] with a < b. Then f(a) = f(b) =| | Since f is continuous
v on the interval -2,-1
and differentiable v on the interval (-2,-1), by Rolle's theorem
v 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[-2, -1]. Thus, f cannot have
more than one root in -2,-1.
(Note: where the problem asks you to make a choice select the weakest choice that works in the given context. For example "continuous" is a
weaker condition than "polynomial" because every polynomial is continuous but not vice-versa. Rolle's theorem is a weaker theorem than the
mean value theorem because Rolle's theorem applies to fewer cases.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8d5a518e-0485-4e1c-badc-2c679a224bb7%2F24c69485-10be-4f7d-83ca-35fb64cb79b8%2Fgj75294_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Suppose f(x) = =°+ 3z + 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
v function on the interval [-2, -1] and f(-2) =
and f(-1) =|
the graph of y = f(x) must cross the r-axis at some point in the interval (-2, –1) by the
intermediate value theorem v
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 experiment. Suppose that there were two
roots z = a and z = b in the interval [-2, -1] with a < b. Then f(a) = f(b) =| | Since f is continuous
v on the interval -2,-1
and differentiable v on the interval (-2,-1), by Rolle's theorem
v 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[-2, -1]. Thus, f cannot have
more than one root in -2,-1.
(Note: where the problem asks you to make a choice select the weakest choice that works in the given context. For example "continuous" is a
weaker condition than "polynomial" because every polynomial is continuous but not vice-versa. Rolle's theorem is a weaker theorem than the
mean value theorem because Rolle's theorem applies to fewer cases.)
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