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anonymous

  • 5 years ago

Suppose that x=2 is a critical point of f(x)=x^(3)e^(−kx). Find the value of k:

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  1. anonymous
    • 5 years ago
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    You could derive using the product rule (x^3)'(e^-kx)+(x^e)(e^-kx)', and set that equal to 0 and plug 2 in for every x to solve for k.

  2. anonymous
    • 5 years ago
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    so the answer would be?? i really need the answer now :/

  3. anonymous
    • 5 years ago
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    3x^2(e^-kx)+(x^3)(-ke^-kx) = 0 3*2^2*(e^-2k)+(2^3)(-ke^-2k) = 0 12e^-2k+-8ke^-2k = 0 e^-2k(12-8k) = 0 ln(e^-2k)+ln(12-8k) = 0 -2k+ln(12-8k) = 0 solve with a calculator? Someone please stop me if I'm doing it wrong. It's late.

  4. anonymous
    • 5 years ago
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    \[12e^{-2k} - 8ke^{-2k} =0 \rightarrow e^{-2k}(12 - 8k) = 0\]

  5. anonymous
    • 5 years ago
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    either e^(-2k) = 0 or 12-8k = 0 => k = 12/8

  6. anonymous
    • 5 years ago
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    other solution is k = 0, I suppose

  7. anonymous
    • 5 years ago
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    Is there any other restrictions? f(2) gets a bigger value when k = 0. I don't know if that's relevant

  8. anonymous
    • 5 years ago
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    no i think 12/8 was the answer . thank you :)

  9. anonymous
    • 5 years ago
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    They both seem correct

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