Transforming Functions: From f(x+1) to f(x)

[Red_CCF]

Hi

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?

Thanks.

 

[Dr. Seafood]

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)

 

[MisterX]

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.

 

[Red_CCF]

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)

Hi

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?

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.

Hi

So f(x+1) = x has the same graph as f (x) = x + 1? Am I allowed to equate the two?

 

[Dr. Seafood]

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?

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).

 

[I_am_learning]

Its really very simple
given f(x+1) = x
so we have
f(2+1) = 2;
f(a+1) = a
Similarly
f((x-1)+1) = x-1
hence, f(x) = x -1;

 

[dimension10]

Hi

Also, what would the graph of f(x+1)=x look like?

Thanks.

A straight line.

http://www.wolframalpha.com/input/?i=Plot+f(x+1)=x

 

[dimension10]

Hi

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))?

f(x)=x-1

 

[pmsrw3]

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?

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.

 

[Red_CCF]

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.

Hi

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?

Thanks

 

[HallsofIvy]

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.

 

[Red_CCF]

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.

Hi

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

Thanks

 

[dimension10]

Hi

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

Thanks

The derivative is the same but the graphs are definitely different.

f(x+1)=x
f(x)=x-1

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

Note that $$\frac{d}{dx}(x-C)=\frac{d}{dx}x$$ 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.

 

[Red_CCF]

The derivative is the same but the graphs are definitely different.

f(x+1)=x
f(x)=x-1

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

Note that $$\frac{d}{dx}(x-C)=\frac{d}{dx}x$$ 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.

Hi

Did you mean f(x) = x - 1 here?

 

[HallsofIvy]

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.

 

[dimension10]

Hi

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

Thanks

I think Wolfram Alpha doesn't get it. It is only taking the right side. Look at the key in this link:

http://www.wolframalpha.com/input/?i=Plot+f(x+1)=x+and+f(x)=x

But the correct one is
http://www.wolframalpha.com/input/?i=Plot+f(x)=x-1+and+f(x)=x

 

[dimension10]

Hi

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?

Thanks

f(x)=x-1 is the only possible solution

 

[Red_CCF]

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.

Hi

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 think Wolfram Alpha doesn't get it. It is only taking the right side. Look at the key in this link:

http://www.wolframalpha.com/input/?i=Plot+f(x+1)=x+and+f(x)=x

But the correct one is
http://www.wolframalpha.com/input/?i=Plot+f(x)=x-1+and+f(x)=x

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?

 

[BloodyFrozen]

Putting the equation in standard terms (f(x)=x-1), you should realize the y-int is -1 or coordinate (0,-1)

 

[Red_CCF]

Putting the equation in standard terms (f(x)=x-1), you should realize the y-int is -1 or coordinate (0,-1)

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

 

[BloodyFrozen]

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

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

 

[Red_CCF]

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

Hi

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?

Thanks

 

[BloodyFrozen]

There still would be a way like it, but it'd take more work to do.

https://www.physicsforums.com/showthread.php?t=518117

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)

 

[Red_CCF]

There still would be a way like it, but it'd take more work to do.

https://www.physicsforums.com/showthread.php?t=518117

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)

Hi

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.

Thanks

 

[dimension10]

Something like f(x+1)=x^2+x+1 is not that hard.

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

 

[HallsofIvy]

Hi

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.

Thanks

First, you should see immediately that the graph is a straight line. Taking x= 0 in f(x+1)= x gives f(1)= 0 so one point in the graph, (x, f(x)), is (1, 0). Taking x= 1 in f(x+1)= x gives f(2)= 1 so another point is (2, 1). Draw the straight line through those two points. You should see that the graph is parallel to the graph of f(x)= x but shifted up 1.

 

[Red_CCF]

First, you should see immediately that the graph is a straight line. Taking x= 0 in f(x+1)= x gives f(1)= 0 so one point in the graph, (x, f(x)), is (1, 0). Taking x= 1 in f(x+1)= x gives f(2)= 1 so another point is (2, 1). Draw the straight line through those two points. You should see that the graph is parallel to the graph of f(x)= x but shifted up 1.

Hi

I was wondering how the graph (x, f(x+1)) or x vs. f(x+1) would be drawn. I keep thinking like this: at x = 0, f (0 + 1) = 0 (as f(x+1)=x) and I get (0,0) as a point but is wrong as it passes through the origin, where is my mistake?

Thank you

 

[BloodyFrozen]

Remeber the function: f(x)?

Well at x = 0, f(0+1)= f(1) ->

f(1) = 0 [from the function f(x+1) = x, we're just plugging in values]

This means when we/you choose x = 1 , the y-value will be 0. So your coordinates are (1,0) not (0,0)

To check if this is right,

f(x) = x-1
f(1) = 1-1
f(1) = 0

Hi

I was wondering how the graph (x, f(x+1)) or x vs. f(x+1) would be drawn. I keep thinking like this: at x = 0, f (0 + 1) = 0 (as f(x+1)=x) and I get (0,0) as a point but is wrong as it passes through the origin, where is my mistake?
Thank you

Your mistake is that your choosing x to be 0, but when you have f(x+1), your actually "altering" the x to become 1 (0+1=1). This is why we would likely write it as f(x)=..., that we'd get the x coordinate we originally wanted (not a different one)

Taking x= 1 in f(x+1)= x gives f(2)= 1 so another point is (2, 1). Draw the straight line through those two points. You should see that the graph is parallel to the graph of f(x)= x but shifted up 1.

Probably clearer from HallsofIvy

Hope this makes sense!

 

[Red_CCF]

Your mistake is that your choosing x to be 0, but when you have f(x+1), your actually "altering" the x to become 1 (0+1=1). This is why we would likely write it as f(x)=..., that we'd get the x coordinate we originally wanted (not a different one)

Hi

In the example you provided, the y-axis of the graph is f(x), if I want the x-axis to be x and y-axis to be f(x+1) the graph would not be the same right? Also, I cannot actually equate f(x) to f(x+1)?

Thanks

 

[BloodyFrozen]

Sorry, but I haven't quite understood what you have said. Perhaps you could clarify or another person could answer. :)

 

[Red_CCF]

Sorry, but I haven't quite understood what you have said. Perhaps you could clarify or another person could answer. :)

Hi

I'm just curious because, f(x+1)=x is f(x)=x-1 shifted one to the left, so their graphs must not look identical right? And so we cannot equate f(x+1)=f(x)?

 

[BloodyFrozen]

Actually, f(x)=x-1 is shifted one down or one to the right (from f(x)=x).

 

[Red_CCF]

Actually, f(x)=x-1 is shifted one down or one to the right (from f(x)=x).

Hi

I meant that the graph of f(x+1) = x is the graph of f(x) = x-1 shifted one left.

 

[BloodyFrozen]

Hi

I meant that the graph of f(x+1) = x is the graph of f(x) = x-1 shifted one left.

They're the same thing

 

[Red_CCF]

They're the same thing

I apologize for dwelling on this, but I'm really confused. I read from Stewart's Calculus that y = f(x+c) shift the graph of y = f(x) c units to the left. If the function is shifted their graphs are different so how can be the same?

Thanks