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

6th Edition

ISBN: 9781465208880

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

Publisher: Kendall Hunt Publishing

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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

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Chapter 2 Solutions

Calculus

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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.

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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.

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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