Suppose f(z) = 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 a = a and a = b in the interval -2, –1] with a < b. Then f(a) = f(b) = Since f is continuous v on the interval -2,-1] v on the interval (-2, -1), by intermediate value theorem v there would exist a point c in interval (a, b) so that and continuous f'(c) = 0. However, the only solution to f'(x) = 0 is = 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.)
Suppose f(z) = 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 a = a and a = b in the interval -2, –1] with a < b. Then f(a) = f(b) = Since f is continuous v on the interval -2,-1] v on the interval (-2, -1), by intermediate value theorem v there would exist a point c in interval (a, b) so that and continuous f'(c) = 0. However, the only solution to f'(x) = 0 is = 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.)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
Problem 99E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...
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![Suppose f(z) = 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 a = a and a = b in the interval -2, -1] with a < b. Then f(a) = f(b) =
Since f is continuous
v on the interval -2, -1
v on the interval (-2, -1), by intermediate value theorem v there would exist a point c in interval (a, b) so that
and continuous
f'(c) = 0. However, the only solution to f'(x) = 0 is a =
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%2Fc6dd3af6-ded2-4953-96ce-ed700b9ac4c2%2F00ywwkt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Suppose f(z) = 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 a = a and a = b in the interval -2, -1] with a < b. Then f(a) = f(b) =
Since f is continuous
v on the interval -2, -1
v on the interval (-2, -1), by intermediate value theorem v there would exist a point c in interval (a, b) so that
and continuous
f'(c) = 0. However, the only solution to f'(x) = 0 is a =
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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