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mitchelsewbaran Group Title

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)

  • one year ago
  • one year ago

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  1. ZeHanz Group Title
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    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?

    • one year ago
  2. mitchelsewbaran Group Title
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    no. :(

    • one year ago
  3. ZeHanz Group Title
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    What is the definition of the 3rd root of a number?

    • one year ago
  4. mitchelsewbaran Group Title
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    The number whose cube is equal to a given number

    • one year ago
  5. ZeHanz Group Title
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    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.

    • one year ago
  6. mitchelsewbaran Group Title
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    so that will get the expression out from under the radical

    • one year ago
  7. mitchelsewbaran Group Title
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    so i'll get x-4+4

    • one year ago
  8. mitchelsewbaran Group Title
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    or x-8

    • one year ago
  9. mitchelsewbaran Group Title
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    @ZeHanz am i correct?

    • one year ago
  10. ZeHanz Group Title
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    You get x-4+4=x in my book...

    • one year ago
  11. mitchelsewbaran Group Title
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    oh srry

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

    • one year ago
  13. mitchelsewbaran Group Title
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    ok. thnx. but where did u get p from?

    • one year ago
  14. ZeHanz Group Title
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    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!

    • one year ago
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