Answer:
(b) (ln(3)/1.3, ∞)
Explanation:
You want the interval on which the rate of change of the logistic function f(x) = 24/(1 +e^(-1.3x)) is decreasing.
Logistic curve
The function f(x) graphs as a curve that increases at an increasing rate until it reaches half its final value, then decreases at a decreasing rate so that it asymptotically approaches its final value.
This transition in rate of change occurs when the exponential term in the denominator becomes equal to the constant term. That is, the rate of growth begins to decrease when ...
1 = 3e^(-1.3x)
0 = ln(3) -1.3x . . . . . take natural logs of both sides
x = ln(3)/1.3 . . . . . . solve for x
The growth rate is decreasing on the interval (ln(3)/1.3, ∞), choice B.
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Additional comment
We could look at where the second derivative is negative.
f'(x) = -24(1 +3e^(-1.3x))^-2·(3e^(-1.3x))(-1.3)
f'(x) = 93.6(e^(1.3x) +6 +9e^(-1.3x))^-1 . . . . expand denominator, multiply numerator and denominator by e^(1.3x)
The second derivative is then ...
f''(x) = -93.6(e^(1.3x) +6 +9e^(-1.3x))^-2·(1.3e^(1.3x) -9(1.3e^(-1.3x)))
The sign matches that of the numerator:
121.68(9e^(-1.3x) -e(1.3x)) < 0
9-e^(2.6x) < 0 . . . . . . multiply by e^(1.3x)/121.68
ln(9) < 2.6x . . . . . . natural logs
x > ln(3)/1.3 . . . . divide by 2.6 and cancel factor of 2
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