inverse functions

[Sarahisme]

hey can i just check if i have done this problem right....

 

For which values of the constants a, b and c is the function f(x) = (x − a)/(bx − c) self inverse?

 

i get

 

for c = 1

a = any real number

and

b = any real number

 

how'd i do?

 

Thanks

 

Sarah

[Sarahisme]

lol, guess theres not too many fans of inverse functions here then

[Algebracus]

[MATH]f(f(x))=\frac{f(x)-a}{bf(x)-c}=\frac{\frac{x-a}{bx-c}-a}{b\frac{x-a}{bx-c}-c} = 1[/MATH] for all x (except those few that does not define f(x) or f(f(x))) if and only if [MATH]\frac{x-a}{bx-c}-a=b\frac{x-a}{bx-c}-c[/MATH] for all x, that is, [MATH]x - a - abx + ac = bx - ba - cbx + c^2[/MATH], that is, [MATH]x(1 - ab + bc - b) = c^2 + a - ac - ba[/MATH]. If this is going to work for every x, then we need both [MATH]1 - ab + bc - b = 0[/MATH] and [MATH]c^2 + a - ac - bc = 0[/MATH]. By checking, we find that [MATH]b=1, a=c[/MATH] is a solution to these equations. We can now look for other solutions, and begin by solving the second equation for b: c = 0 or [MATH]b=c-a+a/c[/MATH]. The first eqation gives a+1=c or [MATH]b=1/(a+1-c)[/MATH]. By this, we check the cases:

 

c = 0 gives 1 = b(a + 1) and a = 0, that is, b = 1.

c = a + 1 gives 1 - ab + ab + b - b = 1 = 0, which cannot be true.

[MATH]b = c - a + a/c = 1/(a + 1 - c)[/MATH] gives [MATH]c = (a + 1 - c)(c^2 - ac + a)[/MATH], or [MATH]c^3 - (2a + 1)c^2 + (a+1)^2c - (a^2+a) = 0[/MATH]. We already know the solution c = a, so we can reduce this to [MATH]c^2 + (a+1)c + (a + 1) = 0[/MATH] by dividing with c - a. This new equation has the solutions [MATH]c_1(a), c_2(a)[/MATH], and from this we can find all triplets (a,b,c) so that f(x) is self-inverse. An example is for instance (3, -6.5, - 2).

 

As a paranthesis, c = 1 gives ab = 1 and 1 = b, that is, a = b = c = 1, a trivial case in which f(x) = 1 for alle x not being 1.

[Sarahisme]

hmm ok so i am not really correct,

 

thanks for that though its greatly appreciated!

[Dapthar]

[MATH]f(f(x))=\frac{f(x)-a}{bf(x)-c}=\frac{\frac{x-a}{bx-c}-a}{b\frac{x-a}{bx-c}-c} = 1[/MATH] for all x (except those few that does not define f(x) or f(f(x))) if and only if [MATH]\frac{x-a}{bx-c}-a=b\frac{x-a}{bx-c}-c[/MATH] for all x, that is, [MATH]x - a - abx + ac = bx - ba - cbx + c^2[/MATH], that is, [MATH]x(1 - ab + bc - b) = c^2 + a - ac - ba[/MATH]. If this is going to work for every x, then we need both [MATH]1 - ab + bc - b = 0[/MATH] and [MATH]c^2 + a - ac - bc = 0[/MATH'].

I don't get why you start looking for specific solutions past this point, since there are an infinite number of them. After you get that the following system must be true:

 

[math]1 - ab + bc - b = 0[/math]

[math]c^2 + a - ac - bc = 0[/math]

 

I would just do the following:

 

[math]1 - ab + bc - b = 0[/math]

[math]\iff 1 + b(-a + c - 1) = 0[/math]

[math]\iff -1 = b(-a + c - 1)[/math]

[math]\iff \frac{-1}{-a + c - 1} = \frac{1}{a - c + 1}=b[/math]

 

Also,

 

[math]c^2 + a - ac - bc = 0[/math]

[math]\iff c^2 + a - ac = bc[/math]

[math]\iff c + \frac{a}{c} - a = b[/math]

 

Thus, since [math]b = b[/math]a solution exists if and only if [math]c + \frac{a}{c} - a = \frac{1}{a - c + 1}[/math]. Of course, if the denominator in that last equality, or the right hand side, or [math]c[/math] is zero, you have to go back and work out a special case.

[Sarahisme]

why should [math]

f(f(x))=\frac{f(x)-a}{bf(x)-c}=\frac{\frac{x-a}{bx-c}-a}{b\frac{x-a}{bx-c}-c} = 1

[/math] why is that not equal to just x?

[Sarahisme]

oops guess that didnt print out correctly, but yeah i think its sort of clear enough? lol maybe...

[Sarahisme]

how many cases are there?

 

i get down to the formula

 

b(x^2) + (c^2)x + a = bc(x^2) + x + ac

 

and then i get the cases (i know there are more cases but these are the only ones i can find so far):

 

Case 1:

c = 1, a = any real number, b = any real number

 

Case 2:

c = 0, a = 0, b = 1/x

 

and then i can't find anymore....but i am sure there must be some more

[Sarahisme]

if anyone is around to help, i sorta need it now lol please...

[matt grime]

We can't say when f is or isn't inverse since you've failed to state the domain and codomain of the function.

 

f isn't even *defined* on the real numbers unless b=0, or unless a/b=c (when the function is identically 1/b).

[Sarahisme]

ok ok, ignore all the above posts....my ones anyways ....and lets start fresh...

 

so the question is:

"For which values of the constants a, b and c is the function f(x) = (x − a)/(bx − c) self inverse?"

 

so far all i can come up with is 2 cases:

 

Case 1)

c = 1

a,b = real numbers

ab != 1

 

(!= means cannot equal, i.e. equals sign with a slash through it)

 

and

 

Case 2)

c = -1

a = 0

b = 0

ab != 1

[matt grime]

And again, you've not stated the domain or codomain rendering the question meaningless.

[Sarahisme]

what do you mean? isnt the domain just values that do not make the denominator 0 in for f(x) and the codomain the range of those values in inverse of f(x)??

 

if not could please please plesae explain further...

[matt grime]

Erm, no, that's not right. f could be a function from R to C_{infty}, it could be a function on any number of domains, to any number of codomains, who says the domain is even a subset of the reals? I know you probably automatically think the domain is implicit, but it isn't. Indeed the most sensible interpretation of your function is as a function from the extended complex numbers (or reals) to itself.

 

I cannot emphasize enough that the domain and codomain of a function are part of the definition of the function. f as written is not a function merely an expression.

 

If you wish to check if the expression f(f(x)=x then do so, but do not state it is "inverse".

 

Sadly this kind of abuse of notation is common, and perpetuated by people who ought to know better, ie even the much vaunted academic undergrad calc texts.

[Sarahisme]

ok ok , i see what your saying and i agree, but lets say then i just want to solve the problem f(f(x)) = x

 

is my two cases right? and is there more cases than that?

 

thanks for you help matty

[matt grime]

I can't say I bothered to solve f(f(x))=x. This is maths: understanding why the method works is more important than actually getting the correct answer.

 

though why in case 2 have you specified a=b=0 AND ab!=1: ab=0 automatically. Nor do I see why you're looking at those two cases in particular.

 

Just solve f(f(x))=x

[Sarahisme]

yes thats what i am saying ...to solve it, there are only certain cases which work.... so yeah can you think of any other cases? or are any of my cases wrong?