Question 1037255
<pre><b>{{{f(x)=x^3+x-1}}}

It's a good idea to graph the equation for f(x):

Get some points: (-2,-11), (-1,-3), (0,-1), (1,1), (2,9)

{{{drawing(400,400,-12,12,-12,12,graph(400,400,-12,12,-12,12,x^3+x-1),
locate(8.5,8.5,y=x),locate(1.6,4.696,"f(x)"),
green(line(-15,-15,15,15)))}}}

That graph passes both the vertical and horizontal line tests.
Therefore it has an inverse which is a function.

The graph of the inverse function will be the reflection of that 
graph in the green line, which is known as the identity line y=x, 
where y and x are identical.

To find the equation of the inverse.

1. Replace f(x) by y

{{{y=x^3+x-1}}}

2.  Interchange x and y

{{{x=y^3+y-1}}}

3. Solve that equation for y.

This involves something pretty advanced.

Write it as

{{{y^3+y-1-x=0}}}

Substitute in Cardano's formula for the solution
to the reduced cubic {{{u^3+a*u+b=0}}}, which is:

{{{u = root(3,-b/2+sqrt((b/2)^2+(a/3)^3))+root(3,-b/2+sqrt((b/2)^2-(a/3)^3))}}}

We substitute y for u, 1 for "a" and (-1-x) for "b".

{{{y = root(3,-(-1-x)/2+sqrt(((-1-x)/2)^2+(1/3)^3))+root(3,-(-1-x)/2+sqrt(((-1-x)/2)^2-(1/3)^3))}}}

Simplify what we can

{{{y = root(3,(1+x)/2+sqrt((1+x)^2/4+1/27))+root(3,(1+x)/4-sqrt((1+x)^2/4+1/27))}}}

Then we change y to f<sup>-1</sup>(x)

{{{"f(x)" = root(3,(1+x)/2+sqrt((1+x)^2/4+1/27))+root(3,(1+x)/4-sqrt((1+x)^2/4+1/27))}}}

We get some points, using the calculator
 (-11,2), (-10,-1.92), (-7,-1.63), (-3,-1), (-1,0), (1,1), (5,1.63),
(8,1.92), (9,2), (10,2.07)

{{{drawing(400,400,-12,12,-12,12,

graph(400,400,-12,12,-12,12,x^3+x-1),
green(line(-15,-15,15,15)),locate(1.6,4.696,"f(x)"),
locate(7,3.5,f^(-1)),locate(8.65,3.1,"(x)"),
locate(8.5,8.5,y=x),

graph(400,400,-12,12,-12,12, (sqrt(3)*sqrt(27x^2+54x+31)+9x+9)^(1/3)/(2^(1/3)3^(2/3))-(2/3)^(1/3)/(sqrt(3)sqrt(27x^2+54x+31)+9x+9)^(1/3) ))}}}

Edwin</pre></b>