Answer :
Certainly! Let's tackle each part of the problem step-by-step.
### Part (a): Graph the function [tex]\( f(x) = \frac{191}{1+4 e^{-3x}} \)[/tex] for [tex]\( x=0 \)[/tex] to [tex]\( x=10 \)[/tex].
To graph the function [tex]\( f(x) = \frac{191}{1+4 e^{-3x}} \)[/tex], we need to plot it over the range [tex]\( x = 0 \)[/tex] to [tex]\( x = 10 \)[/tex].
The function [tex]\( f(x) \)[/tex] is a type of logistic function, which typically starts at some value (close to the initial term) and asymptotically approaches a maximum value. It tends to increase monotonically. Below is the general trend you would expect:
- At [tex]\( x=0 \)[/tex], the function starts at a certain value,
- As [tex]\( x \)[/tex] increases, [tex]\( f(x) \)[/tex] increases and eventually approaches a limiting value.
### Part (b): Find [tex]\( f(0) \)[/tex] and [tex]\( f(10) \)[/tex].
To find [tex]\( f(0) \)[/tex]:
[tex]\[ f(0) = \frac{191}{1 + 4 e^{-3(0)}} = \frac{191}{1 + 4 e^0} = \frac{191}{1 + 4 \cdot 1} = \frac{191}{5} = 38.2 \][/tex]
To find [tex]\( f(10) \)[/tex]:
[tex]\[ f(10) = \frac{191}{1 + 4 e^{-3(10)}} = \frac{191}{1 + 4 e^{-30}} \][/tex]
Since [tex]\( e^{-30} \)[/tex] is a very small number, it can be approximated as zero:
[tex]\[ f(10) \approx \frac{191}{1 + 0} = 191 \][/tex]
### Part (c): Is this function increasing or decreasing?
To determine the behavior of the function, we need to compute its derivative [tex]\( f'(x) \)[/tex] and analyze its sign.
Let's denote the function as:
[tex]\[ f(x) = \frac{191}{1 + 4 e^{-3x}} \][/tex]
Taking the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}\left(\frac{191}{1 + 4 e^{-3x}}\right) \][/tex]
Using the chain rule:
[tex]\[ f'(x) = 191 \cdot \frac{d}{dx}\left(1 + 4 e^{-3x}\right)^{-1} \][/tex]
[tex]\[ f'(x) = 191 \cdot \left(1 + 4 e^{-3x}\right)^{-2} \cdot \left(-4 \cdot (-3) e^{-3x}\right) \][/tex]
[tex]\[ f'(x) = 191 \cdot \left(1 + 4 e^{-3x}\right)^{-2} \cdot 12 e^{-3x} \][/tex]
[tex]\[ f'(x) = \frac{191 \cdot 12 e^{-3x}}{\left(1 + 4 e^{-3x}\right)^2} \][/tex]
Since [tex]\( e^{-3x} \)[/tex] is always positive, [tex]\( f'(x) \)[/tex] is positive for all [tex]\( x \)[/tex]. Therefore, the function [tex]\( f(x) \)[/tex] is increasing.
### Part (d): What is the limiting value of this function?
To determine the limiting value as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{191}{1 + 4 e^{-3x}} \][/tex]
As [tex]\( x \to \infty \)[/tex], [tex]\( e^{-3x} \to 0 \)[/tex]:
[tex]\[ \lim_{x \to \infty} f(x) = \frac{191}{1 + 0} = 191 \][/tex]
The limiting value of the function as [tex]\( x \)[/tex] approaches infinity is 191.
### Graph Options
Given the analysis above, the correct graph should:
1. Start at [tex]\( f(0) = 38.2 \)[/tex],
2. Increase steadily and asymptotically approach [tex]\( f(x) = 191 \)[/tex].
Without seeing the specific graphs labeled A, B, C, and D, I can't choose the exact one, but you would select the graph matching this behavior.
Thus, the answers are:
- Part (b): [tex]\( f(0) = 38.2 \)[/tex], [tex]\( f(10) \approx 191 \)[/tex]
- Part (c): The function is increasing.
- Part (d): The limiting value is 191.
### Part (a): Graph the function [tex]\( f(x) = \frac{191}{1+4 e^{-3x}} \)[/tex] for [tex]\( x=0 \)[/tex] to [tex]\( x=10 \)[/tex].
To graph the function [tex]\( f(x) = \frac{191}{1+4 e^{-3x}} \)[/tex], we need to plot it over the range [tex]\( x = 0 \)[/tex] to [tex]\( x = 10 \)[/tex].
The function [tex]\( f(x) \)[/tex] is a type of logistic function, which typically starts at some value (close to the initial term) and asymptotically approaches a maximum value. It tends to increase monotonically. Below is the general trend you would expect:
- At [tex]\( x=0 \)[/tex], the function starts at a certain value,
- As [tex]\( x \)[/tex] increases, [tex]\( f(x) \)[/tex] increases and eventually approaches a limiting value.
### Part (b): Find [tex]\( f(0) \)[/tex] and [tex]\( f(10) \)[/tex].
To find [tex]\( f(0) \)[/tex]:
[tex]\[ f(0) = \frac{191}{1 + 4 e^{-3(0)}} = \frac{191}{1 + 4 e^0} = \frac{191}{1 + 4 \cdot 1} = \frac{191}{5} = 38.2 \][/tex]
To find [tex]\( f(10) \)[/tex]:
[tex]\[ f(10) = \frac{191}{1 + 4 e^{-3(10)}} = \frac{191}{1 + 4 e^{-30}} \][/tex]
Since [tex]\( e^{-30} \)[/tex] is a very small number, it can be approximated as zero:
[tex]\[ f(10) \approx \frac{191}{1 + 0} = 191 \][/tex]
### Part (c): Is this function increasing or decreasing?
To determine the behavior of the function, we need to compute its derivative [tex]\( f'(x) \)[/tex] and analyze its sign.
Let's denote the function as:
[tex]\[ f(x) = \frac{191}{1 + 4 e^{-3x}} \][/tex]
Taking the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}\left(\frac{191}{1 + 4 e^{-3x}}\right) \][/tex]
Using the chain rule:
[tex]\[ f'(x) = 191 \cdot \frac{d}{dx}\left(1 + 4 e^{-3x}\right)^{-1} \][/tex]
[tex]\[ f'(x) = 191 \cdot \left(1 + 4 e^{-3x}\right)^{-2} \cdot \left(-4 \cdot (-3) e^{-3x}\right) \][/tex]
[tex]\[ f'(x) = 191 \cdot \left(1 + 4 e^{-3x}\right)^{-2} \cdot 12 e^{-3x} \][/tex]
[tex]\[ f'(x) = \frac{191 \cdot 12 e^{-3x}}{\left(1 + 4 e^{-3x}\right)^2} \][/tex]
Since [tex]\( e^{-3x} \)[/tex] is always positive, [tex]\( f'(x) \)[/tex] is positive for all [tex]\( x \)[/tex]. Therefore, the function [tex]\( f(x) \)[/tex] is increasing.
### Part (d): What is the limiting value of this function?
To determine the limiting value as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{191}{1 + 4 e^{-3x}} \][/tex]
As [tex]\( x \to \infty \)[/tex], [tex]\( e^{-3x} \to 0 \)[/tex]:
[tex]\[ \lim_{x \to \infty} f(x) = \frac{191}{1 + 0} = 191 \][/tex]
The limiting value of the function as [tex]\( x \)[/tex] approaches infinity is 191.
### Graph Options
Given the analysis above, the correct graph should:
1. Start at [tex]\( f(0) = 38.2 \)[/tex],
2. Increase steadily and asymptotically approach [tex]\( f(x) = 191 \)[/tex].
Without seeing the specific graphs labeled A, B, C, and D, I can't choose the exact one, but you would select the graph matching this behavior.
Thus, the answers are:
- Part (b): [tex]\( f(0) = 38.2 \)[/tex], [tex]\( f(10) \approx 191 \)[/tex]
- Part (c): The function is increasing.
- Part (d): The limiting value is 191.
