You are asking to find the inverse function of #f(x)# or #f^-1(x)#

When finding the inverse of a function, you are essentially solving for #x# but we could also simply switch the #x# and #y# variables in the equation above and solve for #y# like any other problem such that:

Let #f(x)=y# so...

#y=(x-3)^3+1 -> x=(y-3)^3+1#

Now we can solve for #y#

Subtract #1# from both sides:

#x-1=(y-3)^3+cancel(1-1)#

#x-1=(y-3)^3#

Take the cubed root on each side to cancel out the exponent of #3#

#root(3)(x-1)=root(3)((y-3)^3)#

#root(3)(x-1)=y-3#

Finally, add #3# to both sides

#root(3)(x-1)+3=ycancel(-3+3)#

#root(3)(x-1)+3=y# (This is our inverse function)

I mentioned earlier that finding the inverse of a function means that you are solving for #x# but I also suggested that you could simply just switch #x# and #y# and solve for #y# instead. What I'm going to do now is to show the solution in which we solve for #x# instead of #y#. You'll find that the process is exactly the same with a little tweak at the end:

#y=(x-3)^3+1#

Subtract #1# from both sides:

#y-1=(x-3)^3+cancel(1-1)#

#y-1=(x-3)^3#

Take the cubed root on each side to cancel out the exponent of #3#

#root(3)(y-1)=root(3)((x-3)^3)#

#root(3)(y-1)=x-3#

Finally, add #3# to both sides

#root(3)(y-1)+3=xcancel(-3+3)#

#root(3)(y-1)+3=x#

As you can see, the equation above is almost exactly the same as the other one we solved for except this function is written in terms of #x#. The tweak I was talking about is that you could choose to solve for #x# from the very beginning but you must remember to switch the variables #x# and #y# at the end so that your answer is expressed in terms of #y#. Thus,

#root(3)(y-1)+3=x->root(3)(x-1)+3=y# (This is our inverse function)

We can then let #y# be #f^-1(x)#, the inverse of #f(x)#

#:.f^-1(x)=root(3)(x-1)+3#