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 December 8th, 2016, 05:24 PM #1 Newbie   Joined: Dec 2016 From: United Kingdom Posts: 1 Thanks: 0 Inverse function Hi, I know how to find the inverse of a function, but I don't understand why the method calls for the switch of x and y at the end: so for example we have y=3x+4 my understanding is that the inverse function essentially does the opposite, so instead of the y value being dependent on the x values , the x value will be dependent on the y, so our aim is to get the x on the left hand side y=3x+4 3x=y-4 x=(y-4)/3 I don't really understand why from this point the final answer is f(x)^-1= (x-4)/3 Thank you
 December 8th, 2016, 06:05 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,755 Thanks: 1405 $y=3x+4$ ... if $x=5$, then $y=19$, so the function forms the ordered pair, $(5,19)$ $y=\dfrac{x-4}{3}$ ... if $x=19$, then $y=5$, so this function forms the ordered pair, $(19,5)$ What happened to $(x,y)$ from the first function to the second? Thanks from greg1313 and topsquark
 December 8th, 2016, 06:40 PM #3 Senior Member   Joined: Jul 2016 From: USA Posts: 108 Thanks: 13 Okay I know exactly what you are asking. I wondered this myself too. Basically, at first you have y defined by x. The inverse function defines x with y. Makes sense, just re-arrange the equation where x is defined by y. BEFORE: y=3x+4 AFTER: x=(y-4)/3 So I'm sure you understand this much. But the reason why you should switch the variables at this point is because you want x to be the input and y to be the output. You very well could leave the equation be and let y be your input and x be your output. But we are not used to that and it is hard to understand like that, so that is why we usually switch the two variables at the end, so that y can be the output, rather than the input. Please ask if you have any questions with what I just said! Thanks from topsquark
December 8th, 2016, 09:43 PM   #4
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Quote:
 Originally Posted by z183 I don't understand why the method calls for the switch of x and y at the end:
Maybe your (version of the) method does, but it's not that way in two universities I've tutored
the students in. We interchange x and y at the beginning for our preference.

Quote:
 Originally Posted by VisionaryLen ..., so that is why we > > > usually < < < switch the two variables at the end, so that y can be the output, rather than the input.
You can speak for yourself, but not "usually," unless you have some strong anecdotal evidence.

.

Last edited by Math Message Board tutor; December 8th, 2016 at 09:51 PM.

 December 10th, 2016, 05:36 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,199 Thanks: 873 If function f maps x to y: f(x)= y then its inverse $f^{-1}$ maps y to x: $f^{-1}(y)= x$ If, for example, f(x)= 3x+ 4, equivalent to y= 3x+ 4, then 3x= y- 4 so x= (y- 4)/3. That is, just as f maps x to y= 3x+ 4, $f^{-1}$ maps y to x= (y- 4)/3. Swapping x and y comes from our notational convention of always writing a function in the form "y= f(x)". That is, we want the function writing in terms of x so that instead of writing "$f^{-1}(y)= (y- 4)/3$" we write $f^{-1}(x)= (x- 4)/3$. Those two equations really mean the same thing. Thanks from topsquark
December 10th, 2016, 07:50 PM   #6
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Quote:
 Originally Posted by Country Boy & Math Message Board tutor edit If function f maps x to y: f(x)= y then its inverse $\displaystyle f^{-1}$ maps y to x: $\displaystyle f^{-1}(y)= x$ If, for example, f(x)= 3x+ 4, equivalent to y= 3x+ 4, then 3x= y- 4 so x= (y- 4)/3. That is, just as f maps x to y= 3x+ 4, $\displaystyle f^{-1}$ maps y to x= (y- 4)/3. Swapping x and y comes from our notational convention of always writing a function in the form "y= f(x)". That is, we want the function writing in terms of x so that instead of writing "$\displaystyle f^{-1}(y)= (y- 4)/3$" we write $\displaystyle f^{-1}(x)= (x- 4)/3$. Those two equations really mean the same thing.
Country Boy, in the quote box, I replaced "tex" with "math" and "/tex" with "/math" in your
text so the Latex would work with on this forum.

Please make the necessary adjustments for your Latex posting.

Last edited by Math Message Board tutor; December 10th, 2016 at 07:55 PM.

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