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#