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anonymous

  • one year ago

help on finding inverse functions! f(x) = x^3 - 8

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  1. IrishBoy123
    • one year ago
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    write it as \(y = x^3 - 8\) ad then find \(x = \ ...\) in terms of y

  2. anonymous
    • one year ago
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    i know how to do that, i added 8 to both sides but im stuck on what to do with the cubed x

  3. tkhunny
    • one year ago
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    Have you considered a Cube Root?

  4. anonymous
    • one year ago
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    whats that

  5. IrishBoy123
    • one year ago
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    \(\sqrt[3]{27} = 3\) because \(3 \times 3 \times 3 = 27\)

  6. IrishBoy123
    • one year ago
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    \(\sqrt[3]{whatever}\) is the cube root of whatever

  7. anonymous
    • one year ago
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    oh, so it would be f^-1 (x) = 3√x + 8 with the radical going over both x and 8 or just x?

  8. tkhunny
    • one year ago
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    Pretty much. Or, you can use a 1/3 exponent.

  9. anonymous
    • one year ago
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    its a one-to one function right

  10. tkhunny
    • one year ago
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    In this case, it is.

  11. anonymous
    • one year ago
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    im confused because i know its one of the 2 choices, both have f(-1)x = 3√x + 8, but one has the radicand or whatever its called over the x ONLY, and the other has the radicand over the whole expression (x+8) which one is it?

  12. anonymous
    • one year ago
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  13. tkhunny
    • one year ago
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    You should not be confused. Many exams have fake answers that might be plausible on first inspection. You must pick the right one.

  14. anonymous
    • one year ago
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    its the one i picked

  15. anonymous
    • one year ago
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    with the radicand going over the whole thing

  16. tkhunny
    • one year ago
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    If you took the cube root of the whole thing, then yes. Did you?

  17. anonymous
    • one year ago
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    no only for x^3

  18. tkhunny
    • one year ago
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    \(y = x^{3} - 8\) Swap \(x = y^{3} - 8\) Solve \(x + 8 = y^{3}\) \(\sqrt[3]{x + 8} = y\) Watch yourself work. Don't guess.

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