Question

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

Answer 1

natt888,,are U sure with ur question?

Answer 2

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

Answer 3

is the function have any critical points?

Answer 4

x = 2 is the critical point of the function

Answer 5

\[f(x)=x^3e^{-kx}\rightarrow f'(x)=3x^2 e^{-kx}-kx^3e^{-kx}\]
\[3x^2 e^{-kx}-kx^3e^{-kx}=0\rightarrow \frac {3x^2-kx^3}{e^{kx}}=0\]
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:
\[12-8k=0\rightarrow 12=8k \rightarrow \frac{12}{8}=k \rightarrow k=\frac{3}{2}\]

Answer 6

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.

Answer 7

tianpradana is wrong

Answer 8

whats wrong nadeem?

Answer 9

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.

Answer 10

all you have to do is take the derivative and set the numerator to zero bc the denominator will never be 0