Let f(x)=201+9e−3x .
What is the point of maximum growth rate for the logistic function f(x)=20/1+9e^-3x ?
Round your answer to the nearest hundredth.
(0, 2) <my choice
(5.54, 9)
(0.73, 10)
(0.73, 20)
To find the point of maximum growth rate for the logistic function f(x) = 20/(1 + 9e^(-3x)), we need to determine the derivative of f(x) and find where it is equal to zero.
First, let's find the derivative of f(x). We can use the quotient rule for differentiation:
f'(x) = (20)'(1 + 9e^(-3x)) - 20(1 + 9e^(-3x))' / (1 + 9e^(-3x))^2
The derivative of 20 is 0, and the derivative of (1 + 9e^(-3x)) is -27e^(-3x). Simplifying further:
f'(x) = (-20 * -27e^(-3x)) / (1 + 9e^(-3x))^2
To find where the derivative is equal to zero, set f'(x) = 0 and solve for x:
(-20 * -27e^(-3x)) / (1 + 9e^(-3x))^2 = 0
Since we have a product of two factors, the numerator and the denominator, the only way for the fraction to be equal to zero is if the numerator is equal to zero:
-20 * -27e^(-3x) = 0
Simplifying further:
27e^(-3x) = 0
To solve for x, divide both sides by 27:
e^(-3x) = 0
Since e^(-3x) is a positive value for any real number x, it can never be equal to zero. Therefore, there is no value of x for which the derivative is equal to zero, and consequently, there is no point of maximum growth rate for the logistic function f(x) = 20/(1 + 9e^(-3x)).
Therefore, none of the answer choices provided is correct.
see your previous post.
Why did you repost with the same error in typing?
And are you sure this is algebra 2, and not calculus? Maxima and rates of change are usually not doable with just algebra.