How do you find the inverse of #f(x) = x^3 + 2x# and is it a function?

Let #y = f(x) = x^3+2x#

We want to solve:

#x^3+2x-y = 0#

for #x# (i.e. treating #y# as a constant)

Note that #x^3+2x = x(x^2+2)# is strictly monotonic increasing. So there will be exactly one Real root for any Real value of #y#.

graph{x^3+2x [-10, 10, -5, 5]}

Use Cardano's method...

Let #x = u+v#. Then:

#u^3+v^3+(3uv+2)(u+v)-y=0#

Add the constraint #v=-2/(3u)# to eliminate the #(u+v)# term.

Then our equation becomes:

#u^3-8/(27u^3)-y = 0#

Multiply through by #27u^3# and rearrange slightly to get:

#27(u^3)^2-27y(u^3)-8 = 0#

Using the quadratic formula, we get:

#u^3 = (27y+-sqrt((-27y)^2-4(27)(-8)))/(2*27)#

#=(27y+-sqrt(729y^2+864))/54#

#=(27y+-3sqrt(81y^2+96))/54#

#=(9y+-sqrt(81y^2+96))/18#

Since these roots are Real and the derivation is symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find the Real root of the cubic:

#x = root(3)((9y+sqrt(81y^2+96))/18) + root(3)((9y-sqrt(81y^2+96))/18)#

The other two roots are non-Real complex roots and need not concern us.

This Real zero gives us an inverse function for the original cubic function #f(x)#

#f^(-1)(y) = root(3)((9y+sqrt(81y^2+96))/18) + root(3)((9y-sqrt(81y^2+96))/18)#