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

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

Mathematics
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natt888,,are U sure with ur question?
I mean U're suppose that x=2 is a critical point of f(x)=x^(3)e^(−kx)?
is the function have any critical points?

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x = 2 is the critical point of the function
\[f(x)=x^3e^{-kx}\rightarrow f'(x)=3x^2 e^{-kx}-kx^3e^{-kx}\] \[3x^2 e^{-kx}-kx^3e^{-kx}=0\rightarrow \frac {3x^2-kx^3}{e^{kx}}=0\] 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: \[12-8k=0\rightarrow 12=8k \rightarrow \frac{12}{8}=k \rightarrow k=\frac{3}{2}\]
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.
tianpradana is wrong
whats wrong nadeem?
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.
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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