# Inverse function

1. Oct 10, 2009

### Brisingr24

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. Oct 10, 2009

### g_edgar

Write $y = x^3 + x$, then solve the equation so that it is in the form $x = ...$ with no $x$ 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. Oct 10, 2009

### mathman

What is the question???

4. Oct 10, 2009

### symbolipoint

Brisingr24 wants to know how to find the inverse of f(x)=x3+x.

5. Oct 11, 2009

### HallsofIvy

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. Oct 11, 2009

### rasmhop

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:
$$y^3 + y - x = 0$$
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 $y^2$ 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:
$$y = u+v$$
Now suppose this is the case, then we can substitute this expression into our equation and get:
\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*}
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 $(3uv+1)(u+v)$ 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 $(3uv+1)(u+v)$ 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:
$$u+v=y\qquad 3uv=-1$$
This system has a solution in terms of y and we we will now assume that these two equations are satisfied.
We have:
$$0=u^3 + v^3 + (3uv+1)(u+v) - x = u^3+v^3 -x$$
Multiplying by $u^3$ to get rid of the v's we get:
$$0=u^6+(uv)^3+xu^3 = u^6 + xu^3 - \frac{1}{27}$$
Let $t=u^3$, then we have the quadratic equation:
$$0=t^2+ xt - \frac{1}{27}$$
Which has solution:
$$u^3 = t =\frac{x \pm \sqrt{x^2 + 4/27}}{2}$$
We only need one solution. So we take:
$$u = \sqrt[3]{\frac{x-\sqrt{x^2 + 4/27}}{2}}$$
This gives us the solution:
$$y= u+v =u-\frac{1}{3u} = \sqrt[3]{\frac{x-\sqrt{x^2 + 4/27}}{2}} - \frac{1}{3\sqrt[3]{\frac{x-\sqrt{x^2 + 4/27}}{2}}}$$
So:
$$f^{-1}(x) = \sqrt[3]{\frac{x-\sqrt{x^2 + 4/27}}{2}} - \frac{1}{3\sqrt[3]{\frac{x-\sqrt{x^2 + 4/27}}{2}}}$$
Not exactly pretty and you may be able to simplify it a little bit, but solving cubics isn't pretty in general.

7. Oct 11, 2009

### Brisingr24

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

8. Oct 11, 2009

### Brisingr24

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. Oct 12, 2009

### rasmhop

Alternatively you could simply use the formula for the roots of a depressed cubic (one of the form $x^3 + ax + b$). The equation:
$$y^3 + ay + b = 0$$
Has the solutions:
$$y_1 = \frac{-1}{3}\sqrt[3]{\frac{27b + 3\sqrt{81b^2 + 12a^3}}{2}}-\frac{1}{3}\sqrt[3]{\frac{27b - 3\sqrt{81b^2 + 12a^3}}{2}}$$
$$y_2 = \frac{1+i\sqrt{3}}{6}\sqrt[3]{\frac{27b + 3\sqrt{81b^2 + 12a^3}}{2}}+\frac{1-i\sqrt{3}}{6}\sqrt[3]{\frac{27b - 3\sqrt{81b^2 + 12a^3}}{2}}$$
$$y_3 = \frac{1-i\sqrt{3}}{6}\sqrt[3]{\frac{27b + 3\sqrt{81b^2 + 12a^3}}{2}}+\frac{1+i\sqrt{3}}{6}\sqrt[3]{\frac{27b - 3\sqrt{81b^2 + 12a^3}}{2}}$$
In your specific case we have $a=1$, $b = -x$. So the first solution is:
$$y_1 = \frac{-1}{3}\sqrt[3]{\frac{-27x + 3\sqrt{81x^2 + 12}}{2}} - \frac{1}{3}\sqrt[3]{\frac{-27x - 3\sqrt{81x^2 + 12}}{2}}$$
which works, so an alternative expression for the inverse function is:
$$f^{-1}(x) = \frac{-1}{3}\sqrt[3]{\frac{-27x + 3\sqrt{81x^2 + 12}}{2}} + \frac{1}{3}\sqrt[3]{\frac{27x + 3\sqrt{81x^2 + 12}}{2}}$$

10. Oct 13, 2009

### Brisingr24

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....

11. Oct 14, 2009

### HallsofIvy

It is refered to as "Cardano's cubic formula" and one proof can be found here:

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

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 $b^3= m^3(3^3a^3)$. (Yes, I know $3^3= 27$ but I you will see why I want to leave it like that.)

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

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

Since $n= a^3- b^3$, $b^3= a^3-n$ and so
$$b^3= -\frac{n}{2}\pm \sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}$$

That is the formula rasmhop is using.

Notice that my original equation, $x^3+ mx= n$ has no $x^2$ term. It is what rasmhop called a "depressed cubic". Given a "regular" cubic, $ax^3+ bx^2+ cx+ d$, 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 $y^2$ is 0. Solve that "depressed cubic" for y and use x= y- u to find x.

Last edited by a moderator: May 4, 2017
12. Oct 14, 2009

### Brisingr24

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. Oct 14, 2009

### Hurkyl

Staff Emeritus
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.