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

##### 3 Answers

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:

For 0:

For 2:

Completing g(x) then f(x):

For -2:

For 0:

For 2:

See explanation. Short answer: switch

#### Explanation:

To find the inverse of a function, simply switch the

Let's apply this concept to the first function given and see if we end up with the second function. Also, note

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

#y = x^3 +2#

First switch the variables.

#x = y^3 + 2#

Now we need to solve for

#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

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

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

See explanation...

#### Explanation:

Note that the Real cube root is by definition the inverse of the Real cube function. That is:

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

**Footnote**

The answer above is about Real functions of Real values. The same does not hold if

For example

#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#