Explanation Given information: The given function is g ( t ) = { 3 t + 2 if t ≤ 1 5 if 1 < t ≤ 3 3 t 2 − 1 if t > 3 ...... ( 1 ) Calculation: Since given function has three branches which are polynomial in t , therefore the function is continuous every-where and two points where it is breakable is t = 1 and t = 3 . So, the suspicious points is t = 1 and t = 3 and checking continuity at these two points t = 1 and t = 3 , from (1), lim t → 1 − { g ( t ) } = lim t → 1 − ( 3 t + 2 ) = ( 3 ( 1 ) + 2 ) = 5...... ( 2 ) And, lim t → 1 + { g ( t ) } = lim t → 1 + ( 5 ) = 5...... ( 3 ) Also, g ( 1 ) = ( 3 ( t ) + 2 ) = 5

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Confidence Intervals

When we calculate intervals in probability and statistics, it can be simply termed as finding out a confidence interval. The confidence interval is usually calculated from the data set which is observed. The value of the parameter that needs to be calcul…

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Chapter 2.3, Problem 16PS

To determine

Tofind:All the points of discontinuity for the function

g(t)={3t+2 if t≤15 if 1<t≤33t2−1 if t>3

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Chapter 2.3, Problem 15PS

Chapter 2.3, Problem 17PS

Chapter 2 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

Ch. 2.1 - Prob. 1PS Ch. 2.1 - Prob. 2PS Ch. 2.1 - Prob. 3PS Ch. 2.1 - Prob. 4PS Ch. 2.1 - Prob. 5PS Ch. 2.1 - Prob. 6PS Ch. 2.1 - Prob. 7PS Ch. 2.1 - Prob. 8PS Ch. 2.1 - Prob. 9PS Ch. 2.1 - Prob. 10PS

Ch. 2.1 - Prob. 11PS Ch. 2.1 - Prob. 12PS Ch. 2.1 - Prob. 13PS Ch. 2.1 - Prob. 14PS Ch. 2.1 - Prob. 15PS Ch. 2.1 - Prob. 16PS Ch. 2.1 - Prob. 17PS Ch. 2.1 - Prob. 18PS Ch. 2.1 - Prob. 19PS Ch. 2.1 - Prob. 20PS Ch. 2.1 - Prob. 21PS Ch. 2.1 - Prob. 22PS Ch. 2.1 - Prob. 23PS Ch. 2.1 - Prob. 24PS Ch. 2.1 - Prob. 25PS Ch. 2.1 - Prob. 26PS Ch. 2.1 - Prob. 27PS Ch. 2.1 - Prob. 28PS Ch. 2.1 - Prob. 29PS Ch. 2.1 - Prob. 30PS Ch. 2.1 - Prob. 31PS Ch. 2.1 - Prob. 32PS Ch. 2.1 - Prob. 33PS Ch. 2.1 - Prob. 34PS Ch. 2.1 - Prob. 35PS Ch. 2.1 - Prob. 36PS Ch. 2.1 - Prob. 37PS Ch. 2.1 - Prob. 38PS Ch. 2.1 - Prob. 39PS Ch. 2.1 - Prob. 40PS Ch. 2.1 - Prob. 41PS Ch. 2.1 - Prob. 42PS Ch. 2.1 - Prob. 43PS Ch. 2.1 - Prob. 44PS Ch. 2.1 - Prob. 45PS Ch. 2.1 - Prob. 46PS Ch. 2.1 - Prob. 47PS Ch. 2.1 - Prob. 48PS Ch. 2.1 - Prob. 49PS Ch. 2.1 - Prob. 50PS Ch. 2.1 - Prob. 51PS Ch. 2.1 - Prob. 52PS Ch. 2.1 - Prob. 53PS Ch. 2.1 - Prob. 54PS Ch. 2.1 - Prob. 55PS Ch. 2.1 - Prob. 56PS Ch. 2.1 - Prob. 57PS Ch. 2.1 - Prob. 58PS Ch. 2.1 - 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Prob. 48PS Ch. 2.2 - Prob. 49PS Ch. 2.2 - Prob. 50PS Ch. 2.2 - Prob. 51PS Ch. 2.2 - Prob. 52PS Ch. 2.2 - Prob. 53PS Ch. 2.2 - Prob. 54PS Ch. 2.2 - Prob. 55PS Ch. 2.2 - Prob. 56PS Ch. 2.2 - Prob. 57PS Ch. 2.2 - Prob. 58PS Ch. 2.2 - Prob. 59PS Ch. 2.2 - Prob. 60PS Ch. 2.3 - Prob. 1PS Ch. 2.3 - Prob. 2PS Ch. 2.3 - Prob. 3PS Ch. 2.3 - Prob. 4PS Ch. 2.3 - Prob. 5PS Ch. 2.3 - Prob. 6PS Ch. 2.3 - Prob. 7PS Ch. 2.3 - Prob. 8PS Ch. 2.3 - Prob. 9PS Ch. 2.3 - Prob. 10PS Ch. 2.3 - Prob. 11PS Ch. 2.3 - Prob. 12PS Ch. 2.3 - Prob. 13PS Ch. 2.3 - Prob. 14PS Ch. 2.3 - Prob. 15PS Ch. 2.3 - Prob. 16PS Ch. 2.3 - Prob. 17PS Ch. 2.3 - Prob. 18PS Ch. 2.3 - Prob. 19PS Ch. 2.3 - Prob. 20PS Ch. 2.3 - Prob. 21PS Ch. 2.3 - Prob. 22PS Ch. 2.3 - Prob. 23PS Ch. 2.3 - Prob. 24PS Ch. 2.3 - Prob. 25PS Ch. 2.3 - Prob. 26PS Ch. 2.3 - Prob. 27PS Ch. 2.3 - Prob. 28PS Ch. 2.3 - Prob. 29PS Ch. 2.3 - Prob. 30PS Ch. 2.3 - Prob. 31PS Ch. 2.3 - Prob. 32PS Ch. 2.3 - Prob. 33PS Ch. 2.3 - Prob. 34PS Ch. 2.3 - Prob. 35PS Ch. 2.3 - Prob. 36PS Ch. 2.3 - Prob. 37PS Ch. 2.3 - Prob. 38PS Ch. 2.3 - Prob. 39PS Ch. 2.3 - Prob. 40PS Ch. 2.3 - Prob. 41PS Ch. 2.3 - Prob. 42PS Ch. 2.3 - Prob. 43PS Ch. 2.3 - Prob. 44PS Ch. 2.3 - Prob. 45PS Ch. 2.3 - Prob. 46PS Ch. 2.3 - Prob. 47PS Ch. 2.3 - Prob. 48PS Ch. 2.3 - Prob. 49PS Ch. 2.3 - Prob. 50PS Ch. 2.3 - Prob. 51PS Ch. 2.3 - Prob. 52PS Ch. 2.3 - Prob. 53PS Ch. 2.3 - Prob. 54PS Ch. 2.3 - Prob. 56PS Ch. 2.3 - Prob. 57PS Ch. 2.3 - Prob. 58PS Ch. 2.3 - Prob. 59PS Ch. 2.3 - Prob. 60PS Ch. 2.4 - Prob. 1PS Ch. 2.4 - Prob. 2PS Ch. 2.4 - Prob. 3PS Ch. 2.4 - Prob. 4PS Ch. 2.4 - Prob. 5PS Ch. 2.4 - Prob. 6PS Ch. 2.4 - Prob. 7PS Ch. 2.4 - Prob. 8PS Ch. 2.4 - Prob. 9PS Ch. 2.4 - Prob. 10PS Ch. 2.4 - Prob. 11PS Ch. 2.4 - Prob. 12PS Ch. 2.4 - Prob. 13PS Ch. 2.4 - Prob. 14PS Ch. 2.4 - Prob. 15PS Ch. 2.4 - Prob. 16PS Ch. 2.4 - Prob. 17PS Ch. 2.4 - Prob. 18PS Ch. 2.4 - Prob. 19PS Ch. 2.4 - Prob. 20PS Ch. 2.4 - Prob. 21PS Ch. 2.4 - Prob. 22PS Ch. 2.4 - Prob. 23PS Ch. 2.4 - Prob. 24PS Ch. 2.4 - Prob. 25PS Ch. 2.4 - Prob. 26PS Ch. 2.4 - 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Prob. 17PE Ch. 2 - Prob. 18PE Ch. 2 - Prob. 19PE Ch. 2 - Prob. 20PE Ch. 2 - Prob. 21PE Ch. 2 - Prob. 22PE Ch. 2 - Prob. 23PE Ch. 2 - Prob. 24PE Ch. 2 - Prob. 25PE Ch. 2 - Prob. 26PE Ch. 2 - Prob. 27PE Ch. 2 - Prob. 28PE Ch. 2 - Prob. 29PE Ch. 2 - Prob. 30PE Ch. 2 - Prob. 1SP Ch. 2 - Prob. 2SP Ch. 2 - Prob. 3SP Ch. 2 - Prob. 4SP Ch. 2 - Prob. 5SP Ch. 2 - Prob. 6SP Ch. 2 - Prob. 7SP Ch. 2 - Prob. 8SP Ch. 2 - Prob. 9SP Ch. 2 - Prob. 10SP Ch. 2 - Prob. 11SP Ch. 2 - Prob. 12SP Ch. 2 - Prob. 13SP Ch. 2 - Prob. 14SP Ch. 2 - Prob. 15SP Ch. 2 - Prob. 16SP Ch. 2 - Prob. 17SP Ch. 2 - Prob. 18SP Ch. 2 - Prob. 19SP Ch. 2 - Prob. 20SP Ch. 2 - Prob. 21SP Ch. 2 - Prob. 22SP Ch. 2 - Prob. 23SP Ch. 2 - Prob. 24SP Ch. 2 - Prob. 25SP Ch. 2 - Prob. 26SP Ch. 2 - Prob. 27SP Ch. 2 - Prob. 28SP Ch. 2 - Prob. 29SP Ch. 2 - Prob. 30SP Ch. 2 - Prob. 31SP Ch. 2 - Prob. 32SP Ch. 2 - Prob. 33SP Ch. 2 - Prob. 34SP Ch. 2 - Prob. 35SP Ch. 2 - Prob. 36SP Ch. 2 - Prob. 37SP Ch. 2 - Prob. 38SP Ch. 2 - Prob. 39SP Ch. 2 - Prob. 40SP Ch. 2 - 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Statistics 4.1 Introduction to Inferential Statistics; Author: Dr. Jack L. Jackson II;https://www.youtube.com/watch?v=QLo4TEvBvK4;License: Standard YouTube License, CC-BY

Tofind: All the points of discontinuity for the function g ( t ) = { 3 t + 2 if t ≤ 1 5 if 1 < t ≤ 3 3 t 2 − 1 if t > 3

Solution Summary: The author calculates that the given function is discontinuous at point t=3 and checks continuity at these two points.

Concept explainers

Videos

Tofind:All the points of discontinuity for the function

g(t)={3t+2 if t≤15 if 1<t≤33t2−1 if t>3

Want to see the full answer?

Chapter 2 Solutions

Tofind: All the points of discontinuity for the function g ( t ) = { 3 t + 2 if t ≤ 1 5 if 1 &lt; t ≤ 3 3 t 2 − 1 if t &gt; 3

Solution Summary: The author calculates that the given function is discontinuous at point t=3 and checks continuity at these two points.

Concept explainers

Videos

Tofind:All the points of discontinuity for the function g(t)={3t+2 if t≤15 if 1<t≤33t2−1 if t>3

Want to see the full answer?

Chapter 2 Solutions

Tofind: All the points of discontinuity for the function g ( t ) = { 3 t + 2 if t ≤ 1 5 if 1 < t ≤ 3 3 t 2 − 1 if t > 3

Tofind:All the points of discontinuity for the function

g(t)={3t+2 if t≤15 if 1<t≤33t2−1 if t>3