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Inverse function

  1. Oct 10, 2009 #1
    NOTE: I AM NOT DOING THIS FOR HOMEWORK BUT JUST WANT TO KNOW BECAUSE I COULDN'T HELP SOMEONE ELSE WITH THIS PROBLEM.

    f(x)= x3+x

    can someone please explain how to do this step by step.

    thank you.
     
  2. jcsd
  3. Oct 10, 2009 #2
    Write [itex]y = x^3 + x[/itex], then solve the equation so that it is in the form [itex] x = ... [/itex] with no [itex] x [/itex] on the right-hand side.

    The reason they did not assign this as an exercise in the textbook is that the formulas for solving a cubic equation (as this one is) are far too complicated for low-level math students to deal with.
     
  4. Oct 10, 2009 #3

    mathman

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    What is the question???
     
  5. Oct 10, 2009 #4

    symbolipoint

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    Brisingr24 wants to know how to find the inverse of f(x)=x3+x.
     
  6. Oct 11, 2009 #5

    HallsofIvy

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    In order to find f-1(x) when f(x)= x3+ x, you swap y and x in y= x3+ x and solve for x. It will probably be very difficult to solve that. There is a "cubic formula" but it is complicated.

    Note that this function does have an inverse- its derivative is x2+ 1 which is always positive so it is one-to-one.
     
  7. Oct 11, 2009 #6
    As HallsofIvy said you can see that the function has an inverse since f'(x)>0 for all x. And the inverse is obviously unique so finding a single one suffices. What we really want to do is find a solution to the solution:
    [tex]y^3 + y - x = 0[/tex]
    In terms of y. Now as several posters have already mentioned the general formula is ugly, but you are in a slightly easier case as you have no [itex]y^2[/itex] term and you have concrete coefficients. Just as you can solve a quadratic by completing the square or plugging into a formula, for cubics you can plug into a formula or perform some manipulations to find a solution. One pretty nice way of solving such a cubic solution is known as Cardano's method (you can look it up for the general process). The idea is that instead of having a single variable y we want to introduce two variables u, v that replaces y in such a way that we end up with a quadratic equation. The hard part of this process is realizing that we want:
    [tex]y = u+v[/tex]
    Now suppose this is the case, then we can substitute this expression into our equation and get:
    [tex]\begin{align*}
    (u+v)^3 + (u+v) - x &= u^3 + v^3 + 3u^2v + 3uv^2+u+v - x\\
    &= u^3 + v^3 + 3uv(u + v) + u + v - x \\
    &= u^3 + v^3 + (3uv+1)(u+v) - x = 0
    \end{align*}
    [/tex]
    Now we note that we have only specified one equation which u and v need to satisfy (namely y = u + v), but as they are two unknowns we can impose another one to fully determine them. This remaining equation should let us get rid of [itex](3uv+1)(u+v)[/itex] because if we have both cubed and linear terms we don't have an easier equation than our original one. To get rid of the product [itex](3uv+1)(u+v)[/itex] we can let 3uv+1 =0 or u+v=0, but in the latter case we get a contradiction since 0 = u+v=y then so what we want is variables u,v satisfying:
    [tex]u+v=y\qquad 3uv=-1[/tex]
    This system has a solution in terms of y and we we will now assume that these two equations are satisfied.
    We have:
    [tex]0=u^3 + v^3 + (3uv+1)(u+v) - x = u^3+v^3 -x[/tex]
    Multiplying by [itex]u^3[/itex] to get rid of the v's we get:
    [tex]0=u^6+(uv)^3+xu^3 = u^6 + xu^3 - \frac{1}{27}[/tex]
    Let [itex]t=u^3[/itex], then we have the quadratic equation:
    [tex]0=t^2+ xt - \frac{1}{27}[/tex]
    Which has solution:
    [tex]u^3 = t =\frac{x \pm \sqrt{x^2 + 4/27}}{2}[/tex]
    We only need one solution. So we take:
    [tex]u = \sqrt[3]{\frac{x-\sqrt{x^2 + 4/27}}{2}}[/tex]
    This gives us the solution:
    [tex]y= u+v =u-\frac{1}{3u} = \sqrt[3]{\frac{x-\sqrt{x^2 + 4/27}}{2}} - \frac{1}{3\sqrt[3]{\frac{x-\sqrt{x^2 + 4/27}}{2}}}[/tex]
    So:
    [tex]f^{-1}(x) = \sqrt[3]{\frac{x-\sqrt{x^2 + 4/27}}{2}} - \frac{1}{3\sqrt[3]{\frac{x-\sqrt{x^2 + 4/27}}{2}}}[/tex]
    Not exactly pretty and you may be able to simplify it a little bit, but solving cubics isn't pretty in general.
     
  8. Oct 11, 2009 #7
    Can any one explain how to find the inverse of this funtion
     
  9. Oct 11, 2009 #8
    all i have to say is ...wow... but thank you very much for explaining this... now my challenge is to explain this to someone else.
     
  10. Oct 12, 2009 #9
    Alternatively you could simply use the formula for the roots of a depressed cubic (one of the form [itex]x^3 + ax + b[/itex]). The equation:
    [tex]y^3 + ay + b = 0[/tex]
    Has the solutions:
    [tex]y_1 = \frac{-1}{3}\sqrt[3]{\frac{27b + 3\sqrt{81b^2 + 12a^3}}{2}}-\frac{1}{3}\sqrt[3]{\frac{27b - 3\sqrt{81b^2 + 12a^3}}{2}}[/tex]
    [tex]y_2 = \frac{1+i\sqrt{3}}{6}\sqrt[3]{\frac{27b + 3\sqrt{81b^2 + 12a^3}}{2}}+\frac{1-i\sqrt{3}}{6}\sqrt[3]{\frac{27b - 3\sqrt{81b^2 + 12a^3}}{2}}[/tex]
    [tex]y_3 = \frac{1-i\sqrt{3}}{6}\sqrt[3]{\frac{27b + 3\sqrt{81b^2 + 12a^3}}{2}}+\frac{1+i\sqrt{3}}{6}\sqrt[3]{\frac{27b - 3\sqrt{81b^2 + 12a^3}}{2}}[/tex]
    In your specific case we have [itex]a=1[/itex], [itex]b = -x[/itex]. So the first solution is:
    [tex]y_1 = \frac{-1}{3}\sqrt[3]{\frac{-27x + 3\sqrt{81x^2 + 12}}{2}} - \frac{1}{3}\sqrt[3]{\frac{-27x - 3\sqrt{81x^2 + 12}}{2}}[/tex]
    which works, so an alternative expression for the inverse function is:
    [tex]f^{-1}(x) = \frac{-1}{3}\sqrt[3]{\frac{-27x + 3\sqrt{81x^2 + 12}}{2}} + \frac{1}{3}\sqrt[3]{\frac{27x + 3\sqrt{81x^2 + 12}}{2}}[/tex]
     
  11. Oct 13, 2009 #10
    i dont even know this formula.... the one that im familiar with is the (x^3 plus or minus a number cube) and then u can factor it....
    ihavent seen the form that u have written down..... when do u learn this form.....im in Calculus II and i havent this....

    thanks so much for replying!
     
  12. Oct 14, 2009 #11

    HallsofIvy

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    It is refered to as "Cardano's cubic formula" and one proof can be found here:
    http://www.literka.addr.com/mathcountry/algebra/cubic.htm [Broken]

    But the basic idea is very simple. let x= a- b. Then [itex]x^3= a^3- 3a^2b+ 3ab^2- b^3[/itex]. Also [itex]3abx= 3ab(a- b)= 3a^2b- 3ab^2[/itex]. Adding those [itex]x^3+ 3abx= a^3- b^3[/itex]. In particular, if we let 3ab= m and [itex]a^3- b^3= n[/itex], x= a- b satisifies [itex]x^3+ mx= n[/itex].

    What about the other way around? If we know m and n, can we find a and b and so find x= a- b satisfying the equation? Yes, we can.
    From 3ab= m, b= m/(3a) and so [itex]b^3= m^3(3^3a^3)[/itex]. (Yes, I know [itex]3^3= 27[/itex] but I you will see why I want to leave it like that.)

    Then [itex]n= a^3- b^3= a^3- m^3/(3^3a^3)[/itex]. Multiply both sides of that by [itex]a^3[/itex] to get [itex]na^3= (a^3)^2 m^3/3^3[/itex] so [itex](a^3)^2- na^3+ m^3/3^3[/itex], a quadratic in [itex]a^3[/itex]. By the quadratic formula
    [tex]a^3= \frac{n\pm \sqrt{n^2- 4\frac{m^3}{3^3}}}{2}[/tex]

    Factoring a 2 out of the numerator and canceling the 2 in the denominator gives
    [tex]a^3= \frac{n}{2}\pm \sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}[/tex]

    Since [itex]n= a^3- b^3[/itex], [itex]b^3= a^3-n[/itex] and so
    [tex]b^3= -\frac{n}{2}\pm \sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}[/tex]

    That is the formula rasmhop is using.

    Notice that my original equation, [itex]x^3+ mx= n[/itex] has no [itex]x^2[/itex] term. It is what rasmhop called a "depressed cubic". Given a "regular" cubic, [itex]ax^3+ bx^2+ cx+ d[/itex], you can always define, say, y= x+ u, so that x= y- u, plug that into the equation, and then choose u so that the coefficient of the [itex]y^2[/itex] is 0. Solve that "depressed cubic" for y and use x= y- u to find x.
     
    Last edited by a moderator: May 4, 2017
  13. Oct 14, 2009 #12
    k.. so its a great amount of substitution and a great amont of algebra..... thanks Hall of IVY ans rasmhop and ... this helped me understand it better...

    just another question.... in which course do u learn all of this stuff?

    thanks again guys
     
  14. Oct 14, 2009 #13

    Hurkyl

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    In principle, one can learn very early in their studies that you can work with a function without writing it as an arithmetic expression.


    For nearly any purpose I can think of, the expression "f-1(x)" is much more useful than any equivalent arithmetic expression in x.
     
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