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- anonymous

Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.
f(x) = x^3 + 4 and g(x) = cube root of (x-4)

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- anonymous

- katieb

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- ZeHanz

So, first try f(g(x)), this means put the result of function g in function f:\[f(g(x))=\left( \sqrt[3]{x-4} \right)^3+4=...\]Can you see what the next step is?

- anonymous

no. :(

- ZeHanz

What is the definition of the 3rd root of a number?

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- anonymous

The number whose cube is equal to a given number

- ZeHanz

OK, but not only a given number, but the number from which you wanted to know the 3rd root: e.g: the 3rd root of 8 is 2, because: 2^3=8.
So: the 3rd root of x-4 is p, then p³=x-4. So...\[(\sqrt[3]{x-4})^3=x-4\]by definition.

- anonymous

so that will get the expression out from under the radical

- anonymous

so i'll get x-4+4

- anonymous

or x-8

- anonymous

@ZeHanz am i correct?

- ZeHanz

You get x-4+4=x in my book...

- anonymous

oh srry

- ZeHanz

So we've done the first half of the proof.
Now the second: put the result of f in g:\[g(f(x))=\sqrt[3]{x^3+4-4}\]

- anonymous

ok. thnx. but where did u get p from?

- ZeHanz

p is not important. You can name it anyway you want.
Just look at it this way: say you've got a terribly complex expression, such as:\[3x^{23}-5x^{16}+\frac{ 1 }{ 2x }\]
First, they want you to calculate the 3rd root of it.
I really don't know what it would be, so let's call it p for now:\[p=\sqrt[3]{3x^{23}-5x^{16}+\frac{ 1 }{ 2x }}\]Looks awful, not?
Now you have to calculate .... p³. Well, you don't have to think long about it, youve got your original complex expression back!
The reason for this, that the 3rd root and the 3rd power are each other's inverse.
This means: they undo each other's effect on numbers you put in!

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