Let g(x) = xe^x + be^x, where b is a positive constant
(b) For what positive value of b does g have an absolute maximum at 2/3? Justify your answer
(c) Find all values of b, if any, for which the graph of g has a point of inflection on the interval 0 < x < infinity. Justify your answer.
2 Answers

holy crap i just had this exact free response like a week ago.
Is this a practice one from a year before?
b) take the derivative of the function (Remember product rule!), then set f ‘(x) equal to 0. Then factor out an e^x and solve for b when x equals 2/3.
g'(x) = xe^x + e^x – be^x
0 = xe^x + e^x – be^x
I think the answers 1/3.
c) take the derivative of the derivative, set f ”(x) = 0, then solve for b. dont substitute a number in for b just solve it as a variable.

Derive, set x=2/3 and find where the derivation = 0.
x * e^x ( 1 + b )
udv + vdu
u = x * e^x
du = x * e^x + e^x
v = 1 + b
dv = 0
(1 + b) * (e^x * x + e^x)
(1 + b) * (e^(2/3) * (2/3) + e^(2/3) ) = 0
(1 + b) * e^(2/3) * ( 2/3 + 1) = 0
since e^(2/3) * ( 2/3 + 1) is some nonzero value, then:
1 + b = 0
b = 1
so when b = 1, you have an absolute maximum at x = 2/3
Take the second derivative:
(1 + b) * (e^x * x + e^x)
u = 1+b
du = 0
v = e^x – x * e^x
dv = e^x +x^2 * e^x + e^x = x^2 * e^x
(1 + b) (x^2 * e^x)
(1 + b) (0^2 * e^0) = 0
x = 0, which is where your point of inflection is, no matter what value of b you choose. Since x must be greater than 0, there are no points of inflection in the given domain.