# Continuous Functions

## Homework Statement

The problem is posted below in the picture. I looked at c and d and can do those. I am unsure about a and b.

## The Attempt at a Solution

I looked at graphing the problems, but I think it is a wrong approach.

#### Attachments

• Cont..PNG
12.5 KB · Views: 383

Mark44
Mentor

## Homework Statement

The problem is posted below in the picture. I looked at c and d and can do those. I am unsure about a and b.

## The Attempt at a Solution

I looked at graphing the problems, but I think it is a wrong approach.
Graphing a) and b) might be helpful. What is the definition of continuity at a point?

Graphing a) and b) might be helpful. What is the definition of continuity at a point?
Continuity at a point: A function f is continuous at c if the following three conditions are met.

f(C) is defined

lim as x approaches c exists.

lim as x approaches c of f(x) = f(c)

Mark44
Mentor
You said you were unsure about parts a and b. What are your thoughts about these problems so far?

You said you were unsure about parts a and b. What are your thoughts about these problems so far?
For "a" I was thinking it cant be continuous because the graph would be broken up and you technically could not draw the graph without picking up your pencil.

For "b" I was thinking 0, but I think it is wrong.

Mark44
Mentor
For a) why is |f| continuous? For b) you say, "I was thinking 0." Can you elaborate on this? Try to be as clear as can about what you mean.

In your class has the definition of the limit been given yet? By that I mean the ##\delta - \epsilon## definition?

For a) why is |f| continuous? For b) you say, "I was thinking 0." Can you elaborate on this? Try to be as clear as can about what you mean.

In your class has the definition of the limit been given yet? By that I mean the ##\delta - \epsilon## definition?

Attached is the definition about limits.

I think "a" is not continuous because the graph will be alternating between 1 and -1 and not stay at a straight line. I would have to pick up my pencil to draw the graph.

For "b" I was thinking 0, because the limit as s approaches 0, the graph approaches 0.

#### Attachments

• Limits.PNG
4.5 KB · Views: 371
Mark44
Mentor
Attached is the definition about limits.

I think "a" is not continuous because the graph will be alternating between 1 and -1 and not stay at a straight line.
Also, since you have the limit definition to work with, why isn't f continuous anywhere? Can you say this in some way more rigorous than "picking up the pencil to draw the graph"?
KF33 said:
I would have to pick up my pencil to draw the graph.

For "b" I was thinking 0, because the limit as s approaches 0, the graph approaches 0.
The problem doesn't involve s. Can you write something in more mathematical terms? I.e.,
$$\lim_{x \to 0} f(x) = ?$$
I used LaTeX to write this. In unrendered form this is $$\lim_{x \to 0} f(x) = ?$$

Also, since you have the limit definition to work with, why isn't f continuous anywhere? Can you say this in some way more rigorous than "picking up the pencil to draw the graph"?

The problem doesn't involve s. Can you write something in more mathematical terms? I.e.,
$$\lim_{x \to 0} f(x) = ?$$
I used LaTeX to write this. In unrendered form this is $$\lim_{x \to 0} f(x) = ?$$
For "b" I was thinking 0, because when you look at the limit you see the following.
$$\lim_{x \to 0} f(x) = 0$$
So the function would be defined at that point.

For "a" I was thinking the absolute value function would be continuous, because I am thinking the -1 will change to positive 1.

Mark44
Mentor
For "b" I was thinking 0, because when you look at the limit you see the following.
$$\lim_{x \to 0} f(x) = 0$$
So the function would be defined at that point.
All of the functions listed in parts a through d are defined for all real numbers, so that's not the issue. What are they asking you in this problem?
KF33 said:
For "a" I was thinking the absolute value function would be continuous, because I am thinking the -1 will change to positive 1.
The absolute value of the function, or |f|, is the one that is continuous. If you are given a value of ##\epsilon## would you be able to find a number ##\delta > 0## for which ##| |f(x)| - 0| < \epsilon## when ##|x - 0| < \delta##?

All of the functions listed in parts a through d are defined for all real numbers, so that's not the issue. What are they asking you in this problem?

They are asking me to show f(x) is nowhere continuous. Well it is nowhere continuous because it is not continuous at each point in the interval. I think I got a now. would this work?

Mark44 said:
The absolute value of the function, or |f|, is the one that is continuous. If you are given a value of ##\epsilon## would you be able to find a number ##\delta > 0## for which ##| |f(x)| - 0| < \epsilon## when ##|x - 0| < \delta##?

Is this an idea of what you are getting at.

##\epsilon## = 1/2
##\delta > 0## = 1/4
##| |f(1/8)| - 0| < \epsilon##
when ##|1/8 - 0| < \delta##?