# Inverse functions and One to One

1. Nov 7, 2011

### tmlrlz

1. The problem statement, all variables and given/known data
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

2. Relevant 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.

3. The attempt at a solution
(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.

2. Nov 7, 2011

### lanedance

how is the question actually written?

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

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

then the problem simplifies somewhat by considering f(x)=x

3. Nov 7, 2011

### SammyS

Staff Emeritus
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.
$\displaystyle\frac{ax+b}{cx+d}\quad\quad\quad\frac{-dx+b}{cx-a}\quad \quad=\frac{dx-b}{-cx+a}$​
The second expression is just your f -1 after multiplying the numerator & denominator by -1.

Last edited: Nov 7, 2011
4. Nov 15, 2011

### tmlrlz

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. Nov 15, 2011

### tmlrlz

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. Nov 15, 2011

### SammyS

Staff Emeritus
You can write $f^{-1}(x)$ as $\displaystyle f^{-1}(x)=\frac{-dx+b}{cx-a}\,.$ It's equivalent to what you have even if it looks different.

So you want $f(x)=f^{-1}(x)\,.$

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

$\displaystyle (ax+b)(cx-a)=(-dx+b)(cx+d)$

$\displaystyle acx^2-a^{2}x+bcx-ab=-cdx^2-d^{2}x+bcx+bd$

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. Nov 16, 2011

### tmlrlz

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. Nov 16, 2011

### lanedance

firsxt question was just to calrify, though I think I get it now

here we are considering a map $f : \mathbb{R} \to \mathbb{R}$, and want to find $f^{-1} : \mathbb{R} \to \mathbb{R}$, such that $f^{-1}(x) = f(x), \ \forall x$,

Last edited: Nov 16, 2011
9. Nov 16, 2011

### lanedance

consider first a graph of an arbitrary function $f(x)$, the graph of $f^{-1}(x)$ is given by a reflection in the line y=x. This already limits the possible functions.

Now analytically consider
$$f(x) = f^{-1}(x)$$

Now apply f to both sides
$$f(f(x)) = f(f^{-1}(x))=x$$

2 functions that satisfy this are
$$f(x) = \pm x$$

Last edited: Nov 16, 2011
10. Nov 16, 2011

### lanedance

$$f(x) = \frac{ax+b}{cx+d}$$

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

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

Now equating $f(x)= f^{-1}(x)$ and solving gives
$$f(x) = \frac{ax+b}{cx+d} = \frac{b-dx}{cx-a} = f^{-1}(x)$$
$$\frac{ax+b}{cx+d} = \frac{b-dx}{cx-a}$$
$$(ax+b)(cx-a)= (b-dx)(cx+d)$$
$$acx^2-a^2x+bcx-ab= bcx+bd-dcx^2-d^2x$$
$$x^2c(a+d)+x(d^2-a^2)-b(d+a)=0$$

Last edited: Nov 16, 2011
11. Nov 16, 2011

### lanedance

For this to be true for all x, it gives three conditions to be satisfied:
$c(a+d)=0$, so either $c=0$ or $a=-d$
$d^2-a^2=0$, so either $a=\pm d$
$-b(d+a)=0$, so either $b=0$ or $a=-d$

so going through possible combinations, the valid solutions are
$a=-d$
$c=0$, $a=d$, $b=0$

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: Nov 16, 2011