Given #f(x)=x^3+2# and #g(x)=root3(x-2)# how do you show they are inverses?

3 Answers
Jul 19, 2016

It's not elegant, and I'm sure someone will come along soon with a more-elegant demonstration, but one way would be to substitute in, say, -2, 0 and 2 for x in f(x), find the solution, then substitute that solution into g(x) and see whether it yields the original number. Also the reverse.

Explanation:

Completing f(x) then g(x):

For -2:

#f(-2)=(-2)^3+2=-6#
#g(-6)= root(3)(-6-2) =-2#

For 0:

#f(0)=0^3+2 = 2#
#g(2) = root(3)(2-2)=0#

For 2:

#f(2)=2^3+2=10#
#g(10)=root(3)(10-2)=2#

Completing g(x) then f(x):

For -2:

#g(-2)=root(3)(-2-2)=-1.5874#
#f(-1.5874)=(-1.5874)^3+2=-2#

For 0:

#g(0)=root(3)(0-2)=-1.2599#
#f(-1.2599)=(-1.2599)^3+2=0#

For 2:

#g(2)=root(3)(2-2)=0#
#f(0)=0^3+2=2#

Jul 20, 2016

See explanation. Short answer: switch #x# and #y# and solve for #y#.

Explanation:

To find the inverse of a function, simply switch the #x# and #y# variables and solve for #y#.

Let's apply this concept to the first function given and see if we end up with the second function. Also, note #f(x)# is basically #y#.

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

#y = x^3 +2#

First switch the variables.

#x = y^3 + 2#

Now we need to solve for #y#, so isolate #y#.

#x-2 = y^3 cancel(+2-2)#

#x-2 = y^3#

#root(3)(x-2) = root(3)(y^3)#

#root3(x-2) = y#

We have solved for #y#. Now let's set it equal to the second function and determine if they match.

#root3(x-2) = root(3)(x-2)#

The two functions are identical. Therefore, they are inverses.

Jul 20, 2016

See explanation...

Explanation:

Note that the Real cube root is by definition the inverse of the Real cube function. That is: #root(3)(t^3) = t# and #(root(3)(t))^3 = t# for any Real value of #t#.

So given:

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

#g(x) = root(3)(x-2)#

Then:

#f(g(x)) = (root(3)(x-2))^3+2 = (x-2)+2 = x#

#g(f(x)) = root(3)((x^3+2)-2) = root(3)(x^3) = x#

So #f# and #g# are mutual inverses of one another.

#color(white)()#
Footnote

The answer above is about Real functions of Real values. The same does not hold if #f# and #g# are considered as Complex valued functions of Complex values, essentially because #t -> t^3# is not one to one as a Complex function.

For example #(-1/2+sqrt(3)/2i)^3 = 1#, so we have:

#g(f(-1/2+sqrt(3)/2i))#

#= g((-1/2+sqrt(3)/2i)^3+2)#

#= g(1+2)#

#= g(3)#

#= root(3)(3-2)#

#= root(3)(1) = 1#