Suppose f(z) =z" + 3z +1. Show that f has exactly one root (or zero) in the interval [-6, -1]. Student solution. First, we show that f has a root in the interval (-6, -1). Since f is a choose function on the interval [-6, -1] and f(-6) = and f(-1) = the graph of y = f(z) must cross the z-axis at some point in the interval (-6, -1) by the choose Thus, f has at least one root in the interval -6, -1] Second, we show that f cannot have more than one root in the interval [-6,-1]. Suppose that there were two roots z = a and z = b in the interval [-6, -1] with a

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
icon
Related questions
Question
Suppose f(z) =z'+ 3z +1. Show that f has exactly one root (or zero) in the interval [-6, -1].
Student solution.
First, we show that f has a root in the interval (-6, -1). Since f is a
choose
function on the interval [-6, -1] and
f(-6) = and f(-1) =
the graph of y = f(z) must cross the z-axis at some point in the interval (-6, -1) by the
choose
Thus, f has at least one root in the interval [-6, -1]
Second, we show that f cannot have more than one root in the interval [-6, -1]. Suppose that there were two roots z = a and z = b in the interval [-6, –1] with a < b. Then we have
f(a) = f(b) =
Since f is
choose
on the interval [-6, -1] and
choose
on the interval (-6, -1), by
choose
there would exist a point c in interval (a, b) so that f'(c) = 0. However, the only solution to f'(z) = 0 is
which is not in the interval (a, b), since (a, b) C-6, -1]. Thus, f cannot have more than one root in [-6, -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 weake
theorem than the mean value theorem because Rolle's theorem applies to fewer cases.
Transcribed Image Text:Suppose f(z) =z'+ 3z +1. Show that f has exactly one root (or zero) in the interval [-6, -1]. Student solution. First, we show that f has a root in the interval (-6, -1). Since f is a choose function on the interval [-6, -1] and f(-6) = and f(-1) = the graph of y = f(z) must cross the z-axis at some point in the interval (-6, -1) by the choose Thus, f has at least one root in the interval [-6, -1] Second, we show that f cannot have more than one root in the interval [-6, -1]. Suppose that there were two roots z = a and z = b in the interval [-6, –1] with a < b. Then we have f(a) = f(b) = Since f is choose on the interval [-6, -1] and choose on the interval (-6, -1), by choose there would exist a point c in interval (a, b) so that f'(c) = 0. However, the only solution to f'(z) = 0 is which is not in the interval (a, b), since (a, b) C-6, -1]. Thus, f cannot have more than one root in [-6, -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 weake theorem than the mean value theorem because Rolle's theorem applies to fewer cases.
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

Blurred answer
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Functions and Change: A Modeling Approach to Coll…
Functions and Change: A Modeling Approach to Coll…
Algebra
ISBN:
9781337111348
Author:
Bruce Crauder, Benny Evans, Alan Noell
Publisher:
Cengage Learning