U in the interval (0.5, 1). Part b) Use your graphing calculator to find an interval of length 0.01 that contains a root. Enter your answer as an open interval:
U in the interval (0.5, 1). Part b) Use your graphing calculator to find an interval of length 0.01 that contains a root. Enter your answer as an open interval:
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Do b part
![(When entering numerical values, if necessary round to two decimal places.)
Consider the equation and the interval given below.
In(z+1)-2³ = 0, (0.5,1)
Part a) Use the Intermediate Value Theorem to show that there is a root to the given equation in the
indicated interval.
f(x) = ln(x + 1) - 2³ is
defined continuous positive negative
O
on the closed interval [0.5, 1].
Moreover, f(0.5) = 0.28
Therefore,
f(0.5) <f(1) f(1) < f(0.5)
and f(1)
-0.31
O
Thus, f(0.5) f(1).
Consequently, the conditions of the IVT are met. And, since y = 0 is between f(0.5) and f(1), the IVT
guarantees the existence of a = c in the interval (0.5, 1) such that f(c) = 0.
Therefore, there is a root to the equation ln(x + 1) - ³ - 0 in the interval (0.5, 1).
Part b) Use your graphing calculator to find an interval of length 0.01 that contains a root.
Enter your answer as an open interval:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e2005cb-c51b-49dd-a7c9-1e64cb2e4e1b%2Fdb2b6a79-3b4b-4cf1-b64a-b871e75a92e0%2F4r6by0b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(When entering numerical values, if necessary round to two decimal places.)
Consider the equation and the interval given below.
In(z+1)-2³ = 0, (0.5,1)
Part a) Use the Intermediate Value Theorem to show that there is a root to the given equation in the
indicated interval.
f(x) = ln(x + 1) - 2³ is
defined continuous positive negative
O
on the closed interval [0.5, 1].
Moreover, f(0.5) = 0.28
Therefore,
f(0.5) <f(1) f(1) < f(0.5)
and f(1)
-0.31
O
Thus, f(0.5) f(1).
Consequently, the conditions of the IVT are met. And, since y = 0 is between f(0.5) and f(1), the IVT
guarantees the existence of a = c in the interval (0.5, 1) such that f(c) = 0.
Therefore, there is a root to the equation ln(x + 1) - ³ - 0 in the interval (0.5, 1).
Part b) Use your graphing calculator to find an interval of length 0.01 that contains a root.
Enter your answer as an open interval:
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