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Homework Help: Inverse functions and One to One

  1. Nov 7, 2011 #1
    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
    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.
  2. jcsd
  3. Nov 7, 2011 #2


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    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
  4. Nov 7, 2011 #3


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    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: Nov 7, 2011
  5. Nov 15, 2011 #4
    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?
  6. Nov 15, 2011 #5
    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?
  7. Nov 15, 2011 #6


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    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
  8. Nov 16, 2011 #7
    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?
  9. Nov 16, 2011 #8


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    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: Nov 16, 2011
  10. Nov 16, 2011 #9


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    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: Nov 16, 2011
  11. Nov 16, 2011 #10


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    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: Nov 16, 2011
  12. Nov 16, 2011 #11


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    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

    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
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