f' = e^(-2x) (nx^(n-1) - 2x^n) = x^(n-1) e^(-2x) (n-2x)
f'(3)=0 means (n-6) = 0
so, n=6
If f = x^4 e^(-2x) then
f' = 2x^3 e^(-2x) (2-x)
f" = 4x^2 e^(-2x) (x^2-4x+3)
f"=0 at x=1,3 (and f'≠0)
Consider the function f(x)=x^n e^-2x for x greater than or equal to 0, where n is some unknown constant greater than 2.
A Find the constant n for which the function f(x) attains its maximum value at x=3. Make certain to justify that a maximum is attained. (I got n=6, but idk the steps, pls elaborate and explain)
B. For n=4, find all x values of points of infection for curve f=f(x)
(idk how to do this)
2 answers
I got (1, 1/e^2) and (3, 81/e^6) as my inflection points. Are they right?
1 answer