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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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    natt888,,are U sure with ur question?

  2. anonymous
    • 5 years ago
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    I mean U're suppose that x=2 is a critical point of f(x)=x^(3)e^(−kx)?

  3. anonymous
    • 5 years ago
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    is the function have any critical points?

  4. anonymous
    • 5 years ago
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    x = 2 is the critical point of the function

  5. anonymous
    • 5 years ago
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    \[f(x)=x^3e^{-kx}\rightarrow f'(x)=3x^2 e^{-kx}-kx^3e^{-kx}\] \[3x^2 e^{-kx}-kx^3e^{-kx}=0\rightarrow \frac {3x^2-kx^3}{e^{kx}}=0\] given that x=2 is a critical point plug it in for x to find k: \[\frac {3(2)^2-k(2)^3}{e^{2k}}=0 \rightarrow \frac{12-8k}{e^{2k}}=0\] The denominator cannot be zero for any value of k, so set the numerator to zero: \[12-8k=0\rightarrow 12=8k \rightarrow \frac{12}{8}=k \rightarrow k=\frac{3}{2}\]

  6. anonymous
    • 5 years ago
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    here are my reason.... f(x)=x^(3)e^(−kx) then f'(x) = 3x^2. e^(-kx) + x^3 . e^(-kx)(-k) = 0 This function will never be zero for any real value of x. The exponential is never zero of course and the polynomial will only be zero if x is complex and recall that we only want real values of x for critical points. Therefore, this function will not have any critical points.

  7. anonymous
    • 5 years ago
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    tianpradana is wrong

  8. anonymous
    • 5 years ago
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    whats wrong nadeem?

  9. anonymous
    • 5 years ago
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    It is important to note that not all functions will have critical points,.Do not let this fact lead you to always expect that a function will have critical points.

  10. anonymous
    • 5 years ago
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    all you have to do is take the derivative and set the numerator to zero bc the denominator will never be 0

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