My Math Forum One-to-one Functions

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 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) = ½ 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.
 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).
January 18th, 2014, 02:38 AM   #3
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Re: One-to-one Functions

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 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.

January 18th, 2014, 02:41 AM   #4
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Re: One-to-one Functions

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 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) = ½ 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...

 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 $f(x)= x^{3/5}$ (what you wrote would normally be interpreted as $f(x)= \frac{3}{5}x$ but it is clear from your work that that is not what you meant) the "f(x)= f(y)" becomes $x^{3/5}= y^{3/5}$ so taking the fifth power of both sides, $x^3= y^3$ and then taking the third root of both sides, x= y. Yes, $f(x)= x^{3/5}$ is "one-to-one". Now, had that been $f(x)= x^{2/5}$, there would have been a different answer. Taking the fifth power of both sides would now give $x^2= y^2$ and the square root of both sides is NOT "x= y". The square root of $x^2$ is |x|, not x. Even powers are NOT "one-to-one" while odd powers are. That is because $(-x)^n$, for n and even number, is the same as $x^n$ while for n odd, it is not.
January 18th, 2014, 09:39 AM   #6
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Re: One-to-one Functions

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 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 $f(x)= x^{3/5}$ (what you wrote would normally be interpreted as $f(x)= \frac{3}{5}x$ but it is clear from your work that that is not what you meant) the "f(x)= f(y)" becomes $x^{3/5}= y^{3/5}$ so taking the fifth power of both sides, $x^3= y^3$ and then taking the third root of both sides, x= y. Yes, $f(x)= x^{3/5}$ is "one-to-one". Now, had that been $f(x)= x^{2/5}$, there would have been a different answer. Taking the fifth power of both sides would now give $x^2= y^2$ and the square root of both sides is NOT "x= y". The square root of $x^2$ is |x|, not x. Even powers are NOT "one-to-one" while odd powers are. That is because $(-x)^n$, for n and even number, is the same as $x^n$ while for n odd, it is not.
Thanks. This helps quite a bit.

 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 $f(f^{-1}(x))=x\,(x\in\mathbb{R})$ then f is one-to-one? Or do you need to be more explicit than this?
January 18th, 2014, 11:31 AM   #8
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Re: One-to-one Functions

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 Originally Posted by Slinkey Would I be wrong in thinking that if $f(f^{-1}(x))=x\,(x\in\mathbb{R})$ 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.

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### f={(6,-6), (-1,0),(-5,-6),(-6,1)} one to one function

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