When finding the image of a function, do you need to calculate both the forward and backward (inverse) sets.

More context: the function f(x) maps from R to R, but for certain values of x, it is undefined. So we create a set without those values of x and call it R_x. Then when finding the image of f(R_x), do we need to take into account of the inverse of the function and find what the inverse's equilivant (inputs that are undefined) of x are?

(edited 11 months ago)

Original post by Ignorabimus

More context: the function f(x) maps from R to R, but for certain values of x, it is undefined. So we create a set without those values of x and call it R_x. Then when finding the image of f(R_x), do we need to take into account of the inverse of the function and find what the inverse's equilivant (inputs that are undefined) of x are?

The image is pretty much as you define so loosely f(R_x)

https://en.wikipedia.org/wiki/Image_(mathematics)

There is no need for the inverse to exist, so just think about all the values mapped to by the function.

As mentioned, The image is just whatever the function outputs. It need not be invertible.

For example, the image of the f:R->R, f(x)=0 is the single point set {0}. This f is very not invertible.

As a sidenote, we don't care about invertibility for pre-images either, despite the very unfortunate notation of f^-1(x).

For instance, again take f:R->R, f(x)=0. f^-1(0) = R. So the preimage of a single point could very well be the entire real line.

For example, the image of the f:R->R, f(x)=0 is the single point set {0}. This f is very not invertible.

As a sidenote, we don't care about invertibility for pre-images either, despite the very unfortunate notation of f^-1(x).

For instance, again take f:R->R, f(x)=0. f^-1(0) = R. So the preimage of a single point could very well be the entire real line.

(edited 11 months ago)

Original post by tonyiptony

As mentioned, The image is just whatever the function outputs. It need not be invertible.

For example, the image of the f:R->R, f(x)=0 is the single point set {0}. This f is very not invertible.

As a sidenote, we don't care about invertibility for pre-images either, despite the very unfortunate notation of f^-1(x).

For instance, again take f:R->R, f(x)=0. f^-1(0) = R. So the preimage of a single point could very well be the entire real line.

For example, the image of the f:R->R, f(x)=0 is the single point set {0}. This f is very not invertible.

As a sidenote, we don't care about invertibility for pre-images either, despite the very unfortunate notation of f^-1(x).

For instance, again take f:R->R, f(x)=0. f^-1(0) = R. So the preimage of a single point could very well be the entire real line.

https://postimg.cc/CZgmJR5s

For this question, I found the solution in TSR, but the person who wrote the solution was from several years ago, so I think I won't be able to get in contact. Why would the inverse function's image then? Specifically for the T(S_r) where c != 0 part of finding the image. https://www.thestudentroom.co.uk/showthread.php?t=358891&page=3#post8033440

Original post by Ignorabimus

https://postimg.cc/CZgmJR5s

For this question, I found the solution in TSR, but the person who wrote the solution was from several years ago, so I think I won't be able to get in contact. Why would the inverse function's image then? Specifically for the T(S_r) where c != 0 part of finding the image. https://www.thestudentroom.co.uk/showthread.php?t=358891&page=3#post8033440

For this question, I found the solution in TSR, but the person who wrote the solution was from several years ago, so I think I won't be able to get in contact. Why would the inverse function's image then? Specifically for the T(S_r) where c != 0 part of finding the image. https://www.thestudentroom.co.uk/showthread.php?t=358891&page=3#post8033440

The function excluding that single value is invertible as its effectively 1/x which is invertible apart from x=y=0. To justify the image, they simply draw a parallel with T() which is undefined at a single point and for the inverse map its a/c. You could write T as

T = a/c + (ad/c-b)/(cx-d)

and note that a/c cannot be in the image of T.

(edited 11 months ago)

Original post by mqb2766

The function excluding that single value is invertible as its effectively 1/x which is invertible apart from x=y=0. To justify the image, they simply draw a parallel with T() which is undefined at a single point and for the inverse map its a/c. You could write T as

T = a/c + (ad/c-b)/(cx-d)

and note that a/c cannot be in the image of T.

T = a/c + (ad/c-b)/(cx-d)

and note that a/c cannot be in the image of T.

where does the (ad/c-b) come from, that assumes x=d/c, but r=d/c so its excluded; how is there now an +a/c as T.

So taking the image of S_r, the inverse is taken into account because the inverse is similar?

Sorry, but still not quite understanding that part, I can get the answer, but don't know what the process was and why it had to be done for the answer.

Original post by Ignorabimus

where does the (ad/c-b) come from, that assumes x=d/c, but r=d/c so its excluded; how is there now an +a/c as T.

So taking the image of S_r, the inverse is taken into account because the inverse is similar?

Sorry, but still not quite understanding that part, I can get the answer, but don't know what the process was and why it had to be done for the answer.

So taking the image of S_r, the inverse is taken into account because the inverse is similar?

Sorry, but still not quite understanding that part, I can get the answer, but don't know what the process was and why it had to be done for the answer.

Youre simply dividing the numerator by the denominator and making the fraction proper but you can verify its correct by putting it all over the same denominator cx-d and verify it equals the original improper fraction.

A simplified version is

(x+1)/(x-2) = (x-2+2+1)/(x-2) = 1 + 3/(x-2)

which has a horizontal asymptote at y=1 and a vertical asymptote at x=2. The improper fraction in the question is just a generalised version of this and the image would not include 1 as there is no x such that

3/(x-2) = 0

However it is invertible excluding y=1,x=2 as the function is 1-1.

The question leads you to using the inverse function to define the image because of the symmetry between the function and its inverse. However, you dont need to if you make the fraction proper so there is a horizontal asymptote at a/c, and for a general function the inverse may not exist so you dont need to use it to define the image.

(edited 11 months ago)

Original post by mqb2766

Youre simply dividing the numerator by the denominator and making the fraction proper but you can verify its correct by putting it all over the same denominator cx-d and verify it equals the original improper fraction.

A simplified version is

(x+1)/(x-2) = (x-2+2+1)/(x-2) = 1 + 3/(x-2)

which has a horizontal asymptote at y=1 and a vertical asymptote at x=2. The improper fraction in the question is just a generalised version of this and the image would not include 1 as there is no x such that

3/(x-2) = 0

However it is invertible excluding y=1,x=2 as the function is 1-1.

The question leads you to using the inverse function to define the image because of the symmetry between the function and its inverse. However, you dont need to if you make the fraction proper so there is a horizontal asymptote at a/c, and for a general function the inverse may not exist so you dont need to use it to define the image.

A simplified version is

(x+1)/(x-2) = (x-2+2+1)/(x-2) = 1 + 3/(x-2)

which has a horizontal asymptote at y=1 and a vertical asymptote at x=2. The improper fraction in the question is just a generalised version of this and the image would not include 1 as there is no x such that

3/(x-2) = 0

However it is invertible excluding y=1,x=2 as the function is 1-1.

The question leads you to using the inverse function to define the image because of the symmetry between the function and its inverse. However, you dont need to if you make the fraction proper so there is a horizontal asymptote at a/c, and for a general function the inverse may not exist so you dont need to use it to define the image.

So this is a unique case of images, as in: since the inverse is similar to the original function, we need to take into account of the inverse?

As without taking into account of the inverse, we only get d/c as the undefined, but looking at the inverse, a/c is revealed to be similarly undefinable.

Original post by Ignorabimus

So this is a unique case of images, as in: since the inverse is similar to the original function, we need to take into account of the inverse?

As without taking into account of the inverse, we only get d/c as the undefined, but looking at the inverse, a/c is revealed to be similarly undefinable.

As without taking into account of the inverse, we only get d/c as the undefined, but looking at the inverse, a/c is revealed to be similarly undefinable.

Its a question where theyre hinting at doing it slightly different from normal, though I suspect youd get full marks if you simply did the usual approach as in the previous couple of posts/below. In the question they ask about the domain of T, then get you to show the equivalence of T^-1 and T and then asks about the image of T so ... Its hardly unusual to have a maths question where they ask you about a "simple" concept but do it in an unusual way.

The domain is the real axis without d/c. The image is the real axis without a/c. Its fairly easy to see as

y = ax - b / cx - d

so divide top and bottom by c

y = (a/c x - b/c) / (x - d/c)

Now make the fraction proper

y = a/c + (ad/c^2 - b/c)/(x-d/c)

and take a/c over and denote the new numerator as k = ad/c^2-b/c (non zero by assumption)

y - a/c = k/(x - d/c)

Its just the usual 1/x subject to a horizontal and vertical translation by d/c and a/c and a scaling by k. So a/c and d/c are excluded from the image and domain respectively. This takes no account of the inverse.

(edited 11 months ago)

Original post by mqb2766

Its a question where theyre hinting at doing it slightly different from normal, though I suspect youd get full marks if you simply did the usual approach as in the previous couple of posts/below. In the question they ask about the domain of T, then get you to show the equivalence of T^-1 and T and then asks about the image of T so ... Its hardly unusual to have a maths question where they ask you about a "simple" concept but do it in an unusual way.

The domain is the real axis without d/c. The image is the real axis without a/c. Its fairly easy to see as

y = ax - b / cx - d

so divide top and bottom by c

y = (a/c x - b/c) / (x - d/c)

Now make the fraction proper

y = a/c + (ad/c^2 - b/c)/(x-d/c)

and take a/c over and denote the new numerator as k = ad/c^2-b/c (non zero by assumption)

y - a/c = k/(x - d/c)

Its just the usual 1/x subject to a horizontal and vertical translation by d/c and a/c and a scaling by k. So a/c and d/c are excluded from the image and domain respectively. This takes no account of the inverse.

The domain is the real axis without d/c. The image is the real axis without a/c. Its fairly easy to see as

y = ax - b / cx - d

so divide top and bottom by c

y = (a/c x - b/c) / (x - d/c)

Now make the fraction proper

y = a/c + (ad/c^2 - b/c)/(x-d/c)

and take a/c over and denote the new numerator as k = ad/c^2-b/c (non zero by assumption)

y - a/c = k/(x - d/c)

Its just the usual 1/x subject to a horizontal and vertical translation by d/c and a/c and a scaling by k. So a/c and d/c are excluded from the image and domain respectively. This takes no account of the inverse.

alright, thxs for the detailed answer, the non zero by assumption part was given, so it makes more sense, ad!=bc iirc.

do u know how to close threads, since its been answered so fully?

Original post by Ignorabimus

alright, thxs for the detailed answer, the non zero by assumption part was given, so it makes more sense, ad!=bc iirc.

do u know how to close threads, since its been answered so fully?

do u know how to close threads, since its been answered so fully?

Generally threads only get closed if theyre abused. Its more a case of letting sleeping dogs lie.

One thing I didnt mention when youre thinking about the function and its inverse, its arguably simpler to represent them in the usual basic form

xy = 1

so in this case

(x-d/c)(y-a/c) = k

and all the properties about domain, image, ... can be simply read off as you cant have 0 = k, irrespective of whether the zero is due to the x or y factor (so its largely irrelevant whether youre talking about T or T^-1).

(edited 11 months ago)

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