- #1

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f(x)= x

^{3}+x

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

thank you.

- Thread starter Brisingr24
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- #1

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f(x)= x

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

thank you.

- #2

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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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What is the question???

f(x)= x^{3}+x

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

thank you.

- #4

symbolipoint

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Brisingr24 wants to know how to find the inverse of f(x)=x^{3}+x.

- #5

HallsofIvy

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Note that this function

- #6

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[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

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Can any one explain how to find the inverse of this funtion

f(x)= x^{3}+x

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

thank you.

- #8

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- #9

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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

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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....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]

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.

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

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

thanks again guys

- #13

Hurkyl

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