Remember the definition of an inverse function. If y = f(x) takes you from x to y, then the inverse function x = f-1(y) takes you back from y to x.

So let's find [f of g](x), in other words f(g(x)):

f(g(x)) = f([x^1/3 - 5]/2)

= (2[x^1/3 - 5]/2 + 5)^3

Cancel those 2s:

= (x^1/3 - 5 + 5)^3

Now cancel the 5s:

= (x^1/3)^3

= x

Similarly if we find [g of f](x):

g(f(x)) = ([(2x+5)^3]^1/3 - 5)/2

Cancel the ^3 / ^1/3:

g(f(x)) = (2x+5 - 5)/2

And the 5s:

g(f(x)) = 2x/2

= x

See what's happening? If you start with x, and apply the function f to it, you get f(x). But then if you apply g to that value f(x), you get back the original value x. And the same for g(x) - if you apply thr function f to that, you get back the original x than went in to g.

So the two functions are inverses of each other. If you apply one to x, you can get back the original x by applying the other function to your result.

x -> Apply f -> f(x) -> Apply g -> x

x -> Apply g -> g(x) -> Apply f -> x

Mar 21st, 2015

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