My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Reply
 
LinkBack Thread Tools Display Modes
January 17th, 2014, 07:18 PM   #1
Newbie
 
Joined: Jan 2014

Posts: 19
Thanks: 0

One-to-one Functions

I'm supposed to answer whether or not the function is one-to-one.

1. y = 6 - ?
2. y = 4x - ?
3. f(x) = 4x + 4 - ?
4. g(x) = x + 2 - ?
5. f(x) = x^3/5 - When I take the derivative of this, I get 3/(5x^1/5)^2, which results in a positive number no matter what I plug in for x. So, it is 1-1.
6. g(x) = x^4/5 2 - When I take the derivative of this, I get 4/(5x)^(1/5), which results in a negative number when plugging in a negative number for x and a positive number when plugging in a positive number for x. So, it is not 1-1 because there is a change in sign.

When I take the derivative of problems 1, 2, 3, and 4 I'm left with a whole number. Where do I go from there? I feel like problems 5 and 6 are correct.
ceity is offline  
 
January 17th, 2014, 08:36 PM   #2
Member
 
Joined: Feb 2013

Posts: 80
Thanks: 8

Re: One-to-one Functions

A function is injective (one to one) when each x value corresponds to a unique y value.
That is, much like the vertical line test that they teach you in a high school functions class, you can use a horizontal line test to determine whether or not a function is injective.

Also do note that one can make a function injective by limiting its domain (as frequently done with trigonometric functions).
anakhronizein is offline  
January 18th, 2014, 02:38 AM   #3
Newbie
 
Joined: Jan 2014

Posts: 19
Thanks: 0

Re: One-to-one Functions

Quote:
Originally Posted by anakhronizein
A function is injective (one to one) when each x value corresponds to a unique y value.
That is, much like the vertical line test that they teach you in a high school functions class, you can use a horizontal line test to determine whether or not a function is injective.

Also do note that one can make a function injective by limiting its domain (as frequently done with trigonometric functions).
So, I don't necessarily need to find the derivative - I can just graph the function and do the vertical and horizontal line test? I still don't understand what I need to do after I have the derivative for the first four problems.
ceity is offline  
January 18th, 2014, 02:41 AM   #4
Newbie
 
Joined: Jan 2014

Posts: 19
Thanks: 0

Re: One-to-one Functions

Quote:
Originally Posted by ceity
I'm supposed to answer whether or not the function is one-to-one.

1. y = 6 - ?
2. y = 4x - ?
3. f(x) = 4x + 4 - ?
4. g(x) = x + 2 - ?
5. f(x) = x^3/5 - When I take the derivative of this, I get 3/(5x^1/5)^2, which results in a positive number no matter what I plug in for x. So, it is 1-1.
6. g(x) = x^4/5 2 - When I take the derivative of this, I get 4/(5x)^(1/5), which results in a negative number when plugging in a negative number for x and a positive number when plugging in a positive number for x. So, it is not 1-1 because there is a change in sign.

When I take the derivative of problems 1, 2, 3, and 4 I'm left with a whole number. Where do I go from there? I feel like problems 5 and 6 are correct.
Correction - I meant numbers without variables, not all whole numbers...
ceity is offline  
January 18th, 2014, 04:30 AM   #5
Math Team
 
Joined: Sep 2007

Posts: 2,409
Thanks: 6

Re: One-to-one Functions

Are you saying that you are asked to determine whether or not functions are one-to-one but don't know what "one-to-one" means?

There is no need to take derivatives. "One-to-one" means "if f(x)= f(y) then x= y" so, for example, if f(x)= 6 (a constant) then "f(x)= f(y)" becomes just "6= 6" and tells you nothing about x or y. No, f(x)= 6 is not "one-to-one". If f(x)= 4x then "f(x)= f(y)" becomes 4x= 4y and, dividing both side by 4, x= y. Yes, f(x)= 4x is "one-to-one". If (what you wrote would normally be interpreted as but it is clear from your work that that is not what you meant) the "f(x)= f(y)" becomes so taking the fifth power of both sides, and then taking the third root of both sides, x= y. Yes, is "one-to-one". Now, had that been , there would have been a different answer. Taking the fifth power of both sides would now give and the square root of both sides is NOT "x= y". The square root of is |x|, not x. Even powers are NOT "one-to-one" while odd powers are. That is because , for n and even number, is the same as while for n odd, it is not.
HallsofIvy is offline  
January 18th, 2014, 09:39 AM   #6
Newbie
 
Joined: Jan 2014

Posts: 19
Thanks: 0

Re: One-to-one Functions

Quote:
Originally Posted by HallsofIvy
Are you saying that you are asked to determine whether or not functions are one-to-one but don't know what "one-to-one" means?

There is no need to take derivatives. "One-to-one" means "if f(x)= f(y) then x= y" so, for example, if f(x)= 6 (a constant) then "f(x)= f(y)" becomes just "6= 6" and tells you nothing about x or y. No, f(x)= 6 is not "one-to-one". If f(x)= 4x then "f(x)= f(y)" becomes 4x= 4y and, dividing both side by 4, x= y. Yes, f(x)= 4x is "one-to-one". If (what you wrote would normally be interpreted as but it is clear from your work that that is not what you meant) the "f(x)= f(y)" becomes so taking the fifth power of both sides, and then taking the third root of both sides, x= y. Yes, is "one-to-one". Now, had that been , there would have been a different answer. Taking the fifth power of both sides would now give and the square root of both sides is NOT "x= y". The square root of is |x|, not x. Even powers are NOT "one-to-one" while odd powers are. That is because , for n and even number, is the same as while for n odd, it is not.
Thanks. This helps quite a bit.
ceity is offline  
January 18th, 2014, 11:05 AM   #7
Member
 
Joined: Dec 2013

Posts: 82
Thanks: 0

Re: One-to-one Functions

Would I be wrong in thinking that if then f is one-to-one? Or do you need to be more explicit than this?
Slinkey is offline  
January 18th, 2014, 11:31 AM   #8
Senior Member
 
Joined: Jun 2013
From: London, England

Posts: 1,316
Thanks: 116

Re: One-to-one Functions

Quote:
Originally Posted by Slinkey
Would I be wrong in thinking that if then f is one-to-one? Or do you need to be more explicit than this?
f must be 1-1 for its inverse to be well-defined.

However, if f'(x) > 0 for all x, then f is strictly increasing, hence 1-1.
Pero is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
functions, onetoone



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
functions sheran Algebra 4 July 16th, 2013 07:37 AM
Functions johnny Calculus 2 May 11th, 2011 09:41 PM
Functions Mathmen Algebra 1 November 10th, 2009 06:58 AM
Functions you_of_eh Algebra 4 September 21st, 2009 10:44 AM
Even and Odd Functions PokerKid Algebra 10 October 25th, 2008 08:39 AM





Copyright © 2018 My Math Forum. All rights reserved.