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
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
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