Inverse function

  • Thread starter Brisingr24
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  • #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.
 

Answers and Replies

  • #2
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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.
 
  • #3
mathman
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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.
What is the question???
 
  • #4
symbolipoint
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Brisingr24 wants to know how to find the inverse of f(x)=x3+x.
 
  • #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.
 
  • #6
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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.
 
  • #7
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.
Can any one explain how to find the inverse of this funtion
 
  • #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.
 
  • #9
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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.
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]
 
  • #10
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]
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!
 
  • #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.
 
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  • #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
 
  • #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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