Inverse functions and One to One

  • Thread starter tmlrlz
  • Start date
  • #1
29
0

Homework Statement


For exercises 49 and 50 let f(x) = (ax + b)/(cx + d)
50. Determine the constants a, b, c, d for which f = f-1


Homework Equations


I found in question 49 when they asked to find f-1 that:
f-1 = (dx - b)/(a - cx)
This was also the answer at the back of the book but you are free to double check.


The Attempt at a Solution


I made f(x) = f-1(x)
(ax + b)/(cx + d) = (dx - b)/(a - cx)
a2x - acx2 + ab - bcx = d2x - bcx - bd
From this i realized that finding the values for b and c would be manageable:
For b:
ab + bd = cdx2 + d2x + acx2 - a2x
b ( a + d) = cdx2 + d2x + acx2 - a2x
b = (cdx2 + d2x+ acx2 - a2x) / (a + d)
For c:
-acx - cdx2 = d2x - bd - a2x -ab
c (-ax - dx2) = d2x - bd - a2x -ab
c = (d2x - bd - a2x -ab)/(-ax - dx2)
However i do not know how to find the constants a and d because you cannot just factor them out without leaving them in. Please help, Thank you.
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
how is the question actually written?

note that if the inverse is well defined we have
[tex] f^{-1}((f(x))=x [/tex]

now if the question asks you to find the co-efficients such that it is true for all x that
[tex] f(x)=f^{-1}((f(x))=x [/tex]

then the problem simplifies somewhat by considering f(x)=x
 
  • #3
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,358
1,031

Homework Statement


For exercises 49 and 50 let f(x) = (ax + b)/(cx + d)
50. Determine the constants a, b, c, d for which f = f-1

Homework Equations


I found in question 49 when they asked to find f-1 that:
f-1 = (dx - b)/(a - cx)
This was also the answer at the back of the book but you are free to double check.

The Attempt at a Solution


I made f(x) = f-1(x)
(ax + b)/(cx + d) = (dx - b)/(a - cx)
a2x - acx2 + ab - bcx = d2x - bcx - bd This should be: a2x - acx2 + ab - bcx = cdx2 + d2x - bcx - bd
But, I see you used the correct form below.​

From this i realized that finding the values for b and c would be manageable:
For b:
ab + bd = cdx2 + d2x + acx2 - a2x
b ( a + d) = cdx2 + d2x + acx2 - a2x
b = (cdx2 + d2x+ acx2 - a2x) / (a + d)
For c:
-acx - cdx2 = d2x - bd - a2x -ab
c (-ax - dx2) = d2x - bd - a2x -ab
c = (d2x - bd - a2x -ab)/(-ax - dx2)
However i do not know how to find the constants a and d because you cannot just factor them out without leaving them in. Please help, Thank you.
See correction above.

All you need to do is equate coefficients.
-ac = cd (The coefficients of x2)

a2 - bc = d2 - bc (The coefficients of x)

ab = -bd (The constant terms)​

The solution can be seen visually if you write f(x) and f -1(x) next to each other, as follows.
[itex]\displaystyle\frac{ax+b}{cx+d}\quad\quad\quad\frac{-dx+b}{cx-a}\quad \quad=\frac{dx-b}{-cx+a}[/itex]​
The second expression is just your f -1 after multiplying the numerator & denominator by -1.
 
Last edited:
  • #4
29
0
how is the question actually written?

note that if the inverse is well defined we have
[tex] f^{-1}((f(x))=x [/tex]

now if the question asks you to find the co-efficients such that it is true for all x that
[tex] f(x)=f^{-1}((f(x))=x [/tex]

then the problem simplifies somewhat by considering f(x)=x
The question is for exercises 49 and 50 and i did 49 which is why i did not put it but i think it will help you understand the question better so this is the way the question is written:
For exercises 49 and 50, let f(x) = (ax +b)/(cx +d).
49. a) Show that f is one-to-one iff ad-bc ≠ 0
b) Suppose that ad-bc ≠ 0. Find f-1
50. Determine the constants a,b,c, d for which f = f-1

But i dont understand how i would apply your method here, do you mean that right off the beginning when i make f = f-1 that i should say x = f-1 and work from there?
 
  • #5
29
0
See correction above.

All you need to do is equate coefficients.
-ac = cd (The coefficients of x2)

a2 - bc = d2 - bc (The coefficients of x)

ab = -bd (The constant terms)​

The solution can be seen visually if you write f(x) and f -1(x) next to each other, as follows.
[itex]\displaystyle\frac{ax+b}{cx+d}\quad\quad\quad\frac{-dx+b}{cx-a}\quad \quad=\frac{dx-b}{-cx+a}[/itex]​
The second expression is just your f -1 after multiplying the numerator & denominator by -1.
I'm sorry i dont really understand what you mean, can you elaborate, perhaps with all the variables and values that involve x. I dont understand because for such instances as
ab + bd = acx2 -a2x +cdx2 + d2x
how does that become
ab = bd(acx2 -a2x +cdx2 + d2x)?
How would you equate the coefficients?
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,358
1,031
You can write [itex]f^{-1}(x)[/itex] as [itex]\displaystyle f^{-1}(x)=\frac{-dx+b}{cx-a}\,.[/itex] It's equivalent to what you have even if it looks different.

So you want [itex]f(x)=f^{-1}(x)\,.[/itex]

That gives you: [itex]\displaystyle \frac{ax+b}{cx+d}=\frac{-dx+b}{cx-a}\,.[/itex]

[itex]\displaystyle (ax+b)(cx-a)=(-dx+b)(cx+d)[/itex]

[itex]\displaystyle acx^2-a^{2}x+bcx-ab=-cdx^2-d^{2}x+bcx+bd[/itex]

The coefficient of x2 on the left must be equal to the coefficient of x2 on the right. Similarly the coefficients of x are equal. So are the constant terms.

They ALL give the same result: a=-d
.
 
  • #7
29
0
You can write [itex]f^{-1}(x)[/itex] as [itex]\displaystyle f^{-1}(x)=\frac{-dx+b}{cx-a}\,.[/itex] It's equivalent to what you have even if it looks different.

So you want [itex]f(x)=f^{-1}(x)\,.[/itex]

That gives you: [itex]\displaystyle \frac{ax+b}{cx+d}=\frac{-dx+b}{cx-a}\,.[/itex]

[itex]\displaystyle (ax+b)(cx-a)=(-dx+b)(cx+d)[/itex]

[itex]\displaystyle acx^2-a^{2}x+bcx-ab=-cdx^2-d^{2}x+bcx+bd[/itex]

The coefficient of x2 on the left must be equal to the coefficient of x2 on the right. Similarly the coefficients of x are equal. So are the constant terms.

They ALL give the same result: a=-d
.
Oh i think i get it now, but then what about the values for b and c, are my answers for those still correct or are there no values for b and c because you keep getting a = -d. Is the answer to this question only that a = -d?
 
  • #8
lanedance
Homework Helper
3,304
2
firsxt question was just to calrify, though I think I get it now

here we are considering a map [itex] f : \mathbb{R} \to \mathbb{R}[/itex], and want to find [itex] f^{-1} : \mathbb{R} \to \mathbb{R}[/itex], such that [itex] f^{-1}(x) = f(x), \ \forall x [/itex],
 
Last edited:
  • #9
lanedance
Homework Helper
3,304
2
consider first a graph of an arbitrary function [itex] f(x)[/itex], the graph of [itex] f^{-1}(x) [/itex] is given by a reflection in the line y=x. This already limits the possible functions.

Now analytically consider
[tex]f(x) = f^{-1}(x) [/tex]

Now apply f to both sides
[tex]f(f(x)) = f(f^{-1}(x))=x [/tex]

2 functions that satisfy this are
[tex] f(x) = \pm x [/tex]
 
Last edited:
  • #10
lanedance
Homework Helper
3,304
2
Now consider your function
[tex] f(x) = \frac{ax+b}{cx+d} [/tex]

Now lets follow through the previous working, setting [itex] f(x)= y [/itex] and solving for x gives [itex] f^{-1}(x)[/itex]
[tex] y = \frac{ax+b}{cx+d} [/tex]
[tex] y(cx+d) = ax+b [/tex]
[tex] x(cy-a)= b-dy [/tex]
[tex] x= \frac{b-dy}{cy-a} [/tex]

Hence we have the same as you had fro the inverse (but multiplied by (-1)/(-1)=1)
[tex] f^{-1}(x)= \frac{b-dx}{cx-a} [/tex]

Now equating [itex] f(x)= f^{-1}(x)[/itex] and solving gives
[tex] f(x) = \frac{ax+b}{cx+d} = \frac{b-dx}{cx-a} = f^{-1}(x)[/tex]
[tex] \frac{ax+b}{cx+d} = \frac{b-dx}{cx-a} [/tex]
[tex] (ax+b)(cx-a)= (b-dx)(cx+d) [/tex]
[tex] acx^2-a^2x+bcx-ab= bcx+bd-dcx^2-d^2x [/tex]
[tex] x^2c(a+d)+x(d^2-a^2)-b(d+a)=0 [/tex]
 
Last edited:
  • #11
lanedance
Homework Helper
3,304
2
For this to be true for all x, it gives three conditions to be satisfied:
[itex] c(a+d)=0 [/itex], so either [itex] c=0[/itex] or [itex]a=-d[/itex]
[itex] d^2-a^2=0[/itex], so either [itex] a=\pm d[/itex]
[itex] -b(d+a)=0[/itex], so either [itex] b=0[/itex] or [itex]a=-d[/itex]

so going through possible combinations, the valid solutions are
[itex] a=-d[/itex]
[itex] c=0 [/itex], [itex] a=d[/itex], [itex] b=0[/itex]

the combination
a=1, d=-1, c=0 gives f(x)=x
whilst the combination
a=d, c=0, b=x gives f(x)=-x

now consider the least restrictive a=-d, then
f(x) = \frac{ax+b}{cx-a} = \frac{b+ax}{cx-a} = f^{-1}(x)

Plotting for a=c=b=1
http://www.wolframalpha.com/input/?i=plot+(x+1)/(x-1)

note the symmetry in the y=x line

a=-d solutions represent hyperbolas symmetric in the line y=x (rotated 45degrees), the values of b and c will vary the position and "sharpness" of the hyperbolas
 
Last edited:

Related Threads on Inverse functions and One to One

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
5
Views
1K
Replies
1
Views
889
Replies
5
Views
10K
Top