Question

Demonstrate that f(x) =x^3 and g(x)=cubed square root x are inverses by computing f o g (x) and gof (x).

Answer 1

If you write \(g(x)=\sqrt[3]{x}=x^\frac{1}{3}\), and remember the following rule for powers:

\((a^b)^c=a^{bc}\), you will be safe.

What you have to do, is calculate f(g(x)) and g(f(x)).

I will start with: \(f(g(x))=(\sqrt[3]{x})^3=(x^{\frac{1}{3}})^3=~...(apply~ rule)\)

Then try g(f(x)) yourself!

Answer 2

@ZeHanz what is the rule that is applied though?

Answer 3

It is the rule I wrote about earlier: \((a^b)^c=a^{bc}\)

Answer 4

oh so then it wiould equal to just one?@ZeHanz

Answer 5

Yes, you will get \(x^1=x\) as a result.
If the other calculation gives x as well, you have proved f and g are inverse functions, because they undo each others calculation, so to speak.

Answer 6

so then we just do the same exact thing with g(x)?

Answer 7

@ZeHanz

Answer 8

yes, do g(f(x))

Answer 9

Yes, it has to be both ways to be sure they are each other's inverse.

Answer 10

would it be g(f(x)= cubed square root=x^3 sqaure root? which would equal to x? @ZeHanz

Answer 11

You would get: \(g(f(x))=(x^3)^{\frac{1}{3}}=x^1=x\)

Answer 12

But I think that is what you tried to tell me :)

Answer 13

yeah lol that is but very differentley. :] but what happend to the cubed square root?
@ZeHanz

Answer 14

We changes it to \(x^{\frac{1}{3}}\), remember?

Answer 15

oh okay, and x^3 stays the same ? @ZeHanz

Answer 16

oh okay, and x^3 stays the same ? @ZeHanz

Answer 17

Yes, that is already a good notation for the power rule we applied.