Question

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

Answer 1

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.

Answer 2

so the answer would be?? i really need the answer now :/

Answer 3

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.

Answer 4

\[12e^{-2k} - 8ke^{-2k} =0 \rightarrow e^{-2k}(12 - 8k) = 0\]

Answer 5

either e^(-2k) = 0 or 12-8k = 0 => k = 12/8

Answer 6

other solution is k = 0, I suppose

Answer 7

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

Answer 8

no i think 12/8 was the answer . thank you :)

Answer 9

They both seem correct