Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 3E from Chapter 2.5 from Stewart's Calculus, 8th Edition.
Problem 3E
Chapter:
Problem:
From the graph of f, state the numbers at which f is discontinuous and explain why.
Step-by-Step Solution
Step 1
We are given with a graph of a function. We have to check continuity of the function at certain points.
Step 2: (a)
A function is said to be continuous when its left limit and right limit are equal to function value at that point.
At x=-4 the function value does not exist, so the function is not continuous there
\[f\left( { - 4} \right)\, = \,{\rm{DNE}}\]At x=-2 the the left hand limit of function and right hand limit of function are different, hence the function is not continuous
\[\mathop {{\rm{lim}}\,}\limits_{x = - {2^ - }} \,f\left( x \right) \ne \mathop {{\rm{lim}}\,}\limits_{x = - {2^ + }} \,f\left( x \right)\]At x=2 the the left hand limit of function and right hand limit of function are different, hence the function is not continuous
\[\mathop {{\rm{lim}}\,}\limits_{x = {2^ - }} \,f\left( x \right) \ne \mathop {{\rm{lim}}\,}\limits_{x = {2^ + }} \,f\left( x \right)\]At x=4 the the left hand limit of function and right hand limit of function are different, hence the function is not continuous
\[\mathop {{\rm{lim}}\,}\limits_{x = {4^ - }} \,f\left( x \right) \ne \mathop {{\rm{lim}}\,}\limits_{x = {4^ + }} \,f\left( x \right)\]Therefore, function is discontinuous at
\[x = \left\{ { - 4, - 2,2,4} \right\}\]
Step 3: (b)
At x=-4: as the function value does not exist, it is neither continuous from left nor from the right
At x=-2: as the function continuous from the left but not from the right
At x=2: as the function continuous from the right but not from the left
At x=4: as the function continuous from the right but not from the left
We are given with a graph of a function. We have to check continuity of the function at certain points.
Step 2: (a)
A function is said to be continuous when its left limit and right limit are equal to function value at that point.
At x=-4 the function value does not exist, so the function is not continuous there
\[f\left( { - 4} \right)\, = \,{\rm{DNE}}\]At x=-2 the the left hand limit of function and right hand limit of function are different, hence the function is not continuous
\[\mathop {{\rm{lim}}\,}\limits_{x = - {2^ - }} \,f\left( x \right) \ne \mathop {{\rm{lim}}\,}\limits_{x = - {2^ + }} \,f\left( x \right)\]At x=2 the the left hand limit of function and right hand limit of function are different, hence the function is not continuous
\[\mathop {{\rm{lim}}\,}\limits_{x = {2^ - }} \,f\left( x \right) \ne \mathop {{\rm{lim}}\,}\limits_{x = {2^ + }} \,f\left( x \right)\]At x=4 the the left hand limit of function and right hand limit of function are different, hence the function is not continuous
\[\mathop {{\rm{lim}}\,}\limits_{x = {4^ - }} \,f\left( x \right) \ne \mathop {{\rm{lim}}\,}\limits_{x = {4^ + }} \,f\left( x \right)\]Therefore, function is discontinuous at
\[x = \left\{ { - 4, - 2,2,4} \right\}\]
Step 3: (b)
At x=-4: as the function value does not exist, it is neither continuous from left nor from the right
At x=-2: as the function continuous from the left but not from the right
At x=2: as the function continuous from the right but not from the left
At x=4: as the function continuous from the right but not from the left