To determine the point of maximum growth rate for the logistic function [tex]\( f(x) = \frac{24}{1 + 3 e^{-1.3 x}} \)[/tex], we can consider the inflection point of the function. The inflection point is where the second derivative changes sign, and for a logistic function, it typically occurs at the midpoint of the carrying capacity.
The logistic function [tex]\( f(x) \)[/tex] generally takes the form:
[tex]\[ f(x) = \frac{L}{1 + e^{-k(x-x_0)}} \][/tex]
Here, [tex]\( L \)[/tex] is the carrying capacity of the population, or the maximum value of [tex]\( f(x) \)[/tex]. For the given function:
[tex]\[ L = 24 \][/tex]
The point of maximum growth is at half of the carrying capacity:
[tex]\[ \frac{L}{2} = \frac{24}{2} = 12 \][/tex]
Now we need to find the value of [tex]\( x \)[/tex] where [tex]\( f(x) = 12 \)[/tex]. We solve the equation:
[tex]\[ 12 = \frac{24}{1 + 3 e^{-1.3 x}} \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 12 (1 + 3 e^{-1.3 x}) = 24 \][/tex]
[tex]\[ 1 + 3 e^{-1.3 x} = 2 \][/tex]
[tex]\[ 3 e^{-1.3 x} = 1 \][/tex]
[tex]\[ e^{-1.3 x} = \frac{1}{3} \][/tex]
[tex]\[ -1.3 x = \ln \left( \frac{1}{3} \right) \][/tex]
[tex]\[ -1.3 x = -\ln (3) \][/tex]
[tex]\[ x = \frac{\ln (3)}{1.3} \][/tex]
Using a calculator to find the natural logarithm:
[tex]\[ x \approx \frac{1.0986}{1.3} \approx 0.845 \][/tex]
Rounding to the nearest hundredth:
[tex]\[ x \approx 0.85 \][/tex]
Therefore, the point of maximum growth is:
[tex]\[ (0.85, 12) \][/tex]
Given the options:
- [tex]\((0.85, 12)\)[/tex]
- [tex]\((0.42, 3)\)[/tex]
- [tex]\((0, 4)\)[/tex]
- [tex]\((0.85, 24)\)[/tex]
The correct answer is:
[tex]\[ (0.85, 12) \][/tex]
