# Inverse functions and One to One

• tmlrlz

## 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

(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.

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

## 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

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

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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

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 don't 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?

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.

I'm sorry i don't really understand what you mean, can you elaborate, perhaps with all the variables and values that involve x. I don't 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?

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
.

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
.

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?

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 $f^{-1}(x) = f(x), \ \forall x$,

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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
$$f(x) = \pm x$$

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

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

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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

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