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 2 years ago
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 (x4)
 2 years ago
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 (x4)

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ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0So, first try f(g(x)), this means put the result of function g in function f:\[f(g(x))=\left( \sqrt[3]{x4} \right)^3+4=...\]Can you see what the next step is?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0What is the definition of the 3rd root of a number?

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.0The number whose cube is equal to a given number

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0OK, 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 x4 is p, then p³=x4. So...\[(\sqrt[3]{x4})^3=x4\]by definition.

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.0so that will get the expression out from under the radical

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.0so i'll get x4+4

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.0@ZeHanz am i correct?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0You get x4+4=x in my book...

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0So we've done the first half of the proof. Now the second: put the result of f in g:\[g(f(x))=\sqrt[3]{x^3+44}\]

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.0ok. thnx. but where did u get p from?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0p 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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