Consider the function f(x)=x^n e^(-2x) for x >/= 0, n > 2

Question

A. Find the constant n for which the function f(x) attains its maximum value at x=3.
B. For n=4, find all x values of points of inflection for the curve y=f(x)

Answer 1

f(x)= (x^n)( e^(-2x))
f'(x) = x^n (-2)(e^(-2x) + e^(-2x) (n)x^(n-1)
= 0 when x = 3

e^(-2x) ( -2x^n + nx^(n-1) ) = 0
but x = 3, so
e^-6 (-2(3)^n + n(3)^(n-1) ) = 0
divide out the e^-6
(-2(3)^n + n(3)^(n-1) ) = 0
3^(n-1) (-2(3) + n) = 0
3^(n-1) = 0 ----> so solution
or
-6 + n = 0

n = 6

b)
take the derivative of my f'(x), set it equal to zero, and sub in x = 4
then solve for n
go for it!