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
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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.
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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 non-zero 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.
