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Basic Functions Question

  1. Jul 26, 2011 #1

    I was wondering if it's possible to transform a function like f(x+1) = x to a function that is just in terms of x (f(x))? Also, what would the graph of f(x+1)=x look like?

  2. jcsd
  3. Jul 26, 2011 #2
    It's clear to see what f(x) is if you let y = x + 1. Then x = y - 1 and f(y) = y - 1.
    We can just interchange symbols for convenience of notation and get f(x) = x - 1.

    Another argument is that f is obviously the inverse function of g(x) = x + 1. Thus by inspection (since this particular example is pretty simple), f(x) = x - 1.

    The graph would be a line crossing the vertical axis at (0, -1)
  4. Jul 26, 2011 #3
    f(x + a) is like f(x), but shifted on the x axis by a.

    If one plots the x axis horizontally, with positive x to the right, f(x + a) will look like f(x) shifted to the left by a. If f1(x) = f(x + a), f1(0) = f(a) and f1(1) = f(1 + a). If a is positive, the values of f1(x) are the values of f with a parameter larger than x. Hopefully it is clear to you after reading this why the shifting is to the left by a.
  5. Jul 30, 2011 #4

    So the new x is not the same as the original x? Would this mean that if I were solving a problem and I do this substitution somewhere in the middle the end result would be incorrect as I changed the relationship between the variable and function?


    So f(x+1) = x has the same graph as f (x) = x + 1? Am I allowed to equate the two?
  6. Jul 30, 2011 #5
    The magical part is that you didn't change anything about the function. It's the same function doing the same thing, except now you've made the function's "rule" more explicit by clearly stating f(x). The new x is not the same as the original x, no; but you have no need for the old x because now you have a more useful formula for f(x).
  7. Jul 30, 2011 #6
    Its really very simple
    given f(x+1) = x
    so we have
    f(2+1) = 2;
    f(a+1) = a
    f((x-1)+1) = x-1
    hence, f(x) = x -1;
    Last edited: Jul 30, 2011
  8. Jul 30, 2011 #7
  9. Jul 30, 2011 #8
  10. Jul 30, 2011 #9
    The important thing (and it comes up a lot in mathematics) is that there is nothing special about x in "f(x+1) = x". It's just standing in for any number you want to put in there. That means you can replace the x with anything you like. You could say f(1+1) = 1, or f((a+b)2+1) = (a+b)2, or whatever.

    If you replace x with y-1, as people showed above, you get the particularly useful result f(y) = y-1. Now, in this expression, just as before, there is nothing special about y. It is, once again, just a stand-in for any old number. So now you can replace y with x, giving f(x) = x-1.

    In theory you could have gotten here in one step by replacing x with x-1 in the first expression. But that can be confusing (especially if the expression is more complex), so people seldom do it that way.
  11. Aug 3, 2011 #10


    The thing I'm wondering about is what if the f(x+1) was used in solving an problem which depends on x and its relationship with f. If I substitute it in this way, would I be changing the relationship between the two and the original problem so the resulting solution would be incorrect?

    Also, since graphically f(x+1) = x is the same as f(x) = x how come that's not the answer to my original question?

  12. Aug 3, 2011 #11


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    f(x+ 1)= x is not, graphically or otherwise, the same as f(x)= x. The graph of y= x is a straight line, through the origin, at 45 degrees to the x- axis. f(x+ 1)= x is, as you have been told, the same as f(u)= u- 1 and its graph is a straight line, hrough the point (0, -1), at 45 degrees to the x-axis. The two lines are parallel but not the same.
  13. Aug 3, 2011 #12

    Thanks for the response. I apologize for still not understanding such a simple problem but I just want to be thorough.

    Using the wolfram link dimension10 provided, the graph of f(x+1) = x is a straight line through the origin which is the same as f(x) = x so I'm confused as to how it can be equal to f(u) = u - 1 and not f(x) = x given that f(x+1) and f(u) are not equal for the same x and u values.

    I'm also wondering if someone can answer the other question I posed in Post #10

  14. Aug 4, 2011 #13
    The derivative is the same but the graphs are definitely different.


    f'(x)=d/dx (x-1)

    Note that [tex]\frac{d}{dx}(x-C)=\frac{d}{dx}x[/tex] for constant C.

    As the derivative of f(x)=x is 1, the gradient of the graph is 1. Thus, the angle is arctan 1 =pi/4 which is 45 degrees. So this is the difference is this: if you draw the two graphs, f(x+1)=x and f(x)=x, for every value of f(x), the corresponding points on the two straight lines are 1 unit apart.
  15. Aug 4, 2011 #14

    Did you mean f(x) = x - 1 here?
  16. Aug 4, 2011 #15


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    Science Advisor

    No, he said "f(x+1)= x", which is the same as f(x)= x- 1, and "f(x)= x". Those are the two functions he is comparing.
  17. Aug 5, 2011 #16
    I think Wolfram Alpha doesn't get it. It is only taking the right side. Look at the key in this link:


    But the correct one is
  18. Aug 5, 2011 #17
    f(x)=x-1 is the only possible solution
  19. Aug 5, 2011 #18


    From previous posts, to transform f(x+1) = x to f (x) = x-1, a u = x+1 had to be substituted and then an u = x was substituted back, so essentially an x = x-1 had to be substituted to get f(x) = x-1 so I'm confused how f (x+1) = x can be considered equal to f(x) = x-1 when the x variable has been changed.

    I plotted f(x+1) = x by hand (not sure if it's right) and I got the same graph as the one in the first link. I went f(x+1) = x+1-1; for x = 0, i get f(0+1) = 0+1-1 = 0 (so for x = 0, f = 0) and I get a graph that passes through the origin?
  20. Aug 5, 2011 #19
    Putting the equation in standard terms (f(x)=x-1), you should realize the y-int is -1 or coordinate (0,-1)
  21. Aug 5, 2011 #20
    Hi, I'm not doubting how the graph f(x) = x-1 is drawn, just how it can be equal to f(x+1) = x
  22. Aug 5, 2011 #21
    Think of it this way,

    f(x+1) = x

    We know standard functions are written as f(x) = ...

    We don't know what this f(x) is, but we do know that when we plug x+1 into the function, it yields x.

    Now let g(x) = x+1

    g(x) is the inverse of f(x) or g-1(x) = f(x)-> due to f(g(x)) = x

    To find the inverse of g(x) or in other words f(x) we replace all the y's with x's and x's with y's.

    g(x) = x+1
    x = y+1
    y = x-1 -> g-1(x) = x-1 or what we are looking for, f(x) = x-1

    Although I still think the previous posters' suggestions are much easier than mine:biggrin:
    Last edited: Aug 5, 2011
  23. Aug 5, 2011 #22

    Thanks very much. This makes a lot more sense to me now, I was mainly confused with the other method because of the (essentially) x = x -1 substitution which your method skipped.

    But I wonder, what if f(x+1) was not x, but something else (maybe some polynomial), this method wouldn't work?

  24. Aug 5, 2011 #23
    There still would be a way like it, but it'd take more work to do.


    The thread above has a problem similar to the polynomial/etc. your proposing.

    Basically you have f(g(x)) or f o g [same thing] and g(x) and you need to find f(x).

    I think SammyS's suggestion is a good one to get f(x):wink:
  25. Aug 5, 2011 #24

    I'm wondering if you can show me how to plot f(x+1) = x (without converting to f(x) = x-1) by hand because when I did it I keep getting that the function passes through the origin.

  26. Aug 6, 2011 #25
    Something like f(x+1)=x^2+x+1 is not that hard.

    The solution is f(x)=(x-1)^2+(x-1)+1
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