What is the inverse of f(x) = x^3 + 2x?

tomcenjerrym
Messages
37
Reaction score
0
What is the inverse of this function f(x) = x^3 + 2x?

***************************************

f(x) = x^3 + 2x

y = x^3 + 2x

interchange x \Leftrightarrow y

x = y^3 + 2y

and it's a dead end to me ...
 
Physics news on Phys.org
That will not get you where you need to be. Rather then simply interchanging x and y, you need to chang your expression for f(x) as F(y). A simple example:

f(x)= 3x + 5

y = 3x + 5

y - 5 = 3x

y/3 - 5/3 = x

x = y/3 - 5/3

so now you have x = F(y)
 
What you need to do is solve the equation x3+ 2x= y for x. Thats solving a fairly general cubic equation. Do you have any reason to think that the inverse can be written in a simple form?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top