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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:
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0natt888,,are U sure with ur question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean U're suppose that x=2 is a critical point of f(x)=x^(3)e^(−kx)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the function have any critical points?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = 2 is the critical point of the function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=x^3e^{kx}\rightarrow f'(x)=3x^2 e^{kx}kx^3e^{kx}\] \[3x^2 e^{kx}kx^3e^{kx}=0\rightarrow \frac {3x^2kx^3}{e^{kx}}=0\] given that x=2 is a critical point plug it in for x to find k: \[\frac {3(2)^2k(2)^3}{e^{2k}}=0 \rightarrow \frac{128k}{e^{2k}}=0\] The denominator cannot be zero for any value of k, so set the numerator to zero: \[128k=0\rightarrow 12=8k \rightarrow \frac{12}{8}=k \rightarrow k=\frac{3}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0all 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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