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jesusisabiscuit

  • one year ago

The derivative of f(x)=(x^4/3)-(x^5/5) attains its maximum value at what x value? a. -1 b. 0 c. 1 d. 4/3 e. 5/3

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  1. Lilmoney
    • one year ago
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    B

  2. jesusisabiscuit
    • one year ago
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    can you explain it to me, please?

  3. Lilmoney
    • one year ago
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    Hold on

  4. Yahoo!
    • one year ago
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    find f''(x)

  5. jesusisabiscuit
    • one year ago
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    f''(x) = 4x^2 - 3x^3 what next?

  6. Lilmoney
    • one year ago
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    Isn't that it

  7. wio
    • one year ago
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    Ummm, looks like your \(f''(x)\) is a bit off.

  8. Yahoo!
    • one year ago
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    The condition To attain Max is f''(x) > 0

  9. Lilmoney
    • one year ago
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    That is true

  10. wio
    • one year ago
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    They want the max of the derivative.

  11. wio
    • one year ago
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    So you want to find the critical numbers \(f''(x)=0\) or undefined. Then you'd plug them back into \(f'(x)\) to see where the max is.

  12. Yahoo!
    • one year ago
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    nope @wio they ask..For where the max is attained

  13. wio
    • one year ago
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    "The derivative of f(x)=(x^4/3)-(x^5/5) attains its maximum" "f'(x) attains its maximum"

  14. jesusisabiscuit
    • one year ago
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    aha! f"(x) = 4x^2 - 4x^3. i got my critical numbers. x = 0 & x = 1?

  15. KingGeorge
    • one year ago
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    That looks good. So your local maximum value is at either x=0 or x=1. Also, since your first derivative is a negative quartic (has a \(-x^4\) term), you have a parabola type shape opening downwards, so the local maximum is the global maximum. Now you just need to determine which one is the maximum.

  16. KingGeorge
    • one year ago
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    make sense so far?

  17. jesusisabiscuit
    • one year ago
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    i think. i got f'(0) = 0 and f'(1) = 1/3. so the answer is... 1?

  18. KingGeorge
    • one year ago
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    Bingo.

  19. jesusisabiscuit
    • one year ago
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    All riiiiiiiiight. thank you. and wio. but s/he left.

  20. KingGeorge
    • one year ago
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    You're welcome.

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