# Help on a AP Calculus free response problem?

0

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.

• 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 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.

Also Check This  Kohls orientation question..?