Sure, let's address each part of the question step-by-step.
### a. Graph the function [tex]\( f(x) = \frac{152}{1 + 4e^{-2x}} \)[/tex] for [tex]\( x = 0 \)[/tex] to [tex]\( x = 10 \)[/tex].
To create the graph, we will analyze how the function behaves over the specified range. The function has the form of a logistic function, often used to model growth processes. Since specific graphs are not provided here, let's focus on what the graph should look like:
- At [tex]\( x = 0 \)[/tex]: We can calculate the value directly.
- As [tex]\( x \)[/tex] increases: We should expect the value of [tex]\( f(x) \)[/tex] to increase.
- Long-term behavior: As [tex]\( x \)[/tex] approaches a large positive value, [tex]\( f(x) \)[/tex] reaches a limiting or asymptotic value.
Given this information, the graph should show a smooth curve starting at a certain value at [tex]\( x = 0 \)[/tex] and approaching a higher value as [tex]\( x \)[/tex] goes from 0 to 10.
### b. Find [tex]\( f(0) \)[/tex] and [tex]\( f(10) \)[/tex].
To find [tex]\( f(0) \)[/tex]:
[tex]\[ f(0) = \frac{152}{1 + 4e^{-2 \cdot 0}} = \frac{152}{1 + 4 \cdot 1} = \frac{152}{5} = 30.4 \][/tex]
To find [tex]\( f(10) \)[/tex]:
[tex]\[ f(10) = \frac{152}{1 + 4e^{-2 \cdot 10}} \][/tex]
Since [tex]\( e^{-20} \)[/tex] is exceedingly small ([tex]\( e^{-20} \approx 2.061 \times 10^{-9} \)[/tex]):
[tex]\[ f(10) \approx \frac{152}{1 + 4 \cdot 2.061 \times 10^{-9}} \approx \frac{152}{1} = 152 \][/tex]
### c. Is this function increasing or decreasing?
We need to determine whether [tex]\( f(x) \)[/tex] is increasing or decreasing. By examining the function's form:
[tex]\[ f(x) = \frac{152}{1 + 4e^{-2x}} \][/tex]
- As [tex]\( x \)[/tex] increases, [tex]\( e^{-2x} \)[/tex] decreases.
- Therefore, as [tex]\( x \)[/tex] increases, the denominator [tex]\( 1 + 4e^{-2x} \)[/tex] decreases, which means the overall fraction value [tex]\( f(x) \)[/tex] increases.
Thus, the function [tex]\( f(x) \)[/tex] is increasing as [tex]\( x \)[/tex] increases.
### d. What is the limiting value of this function?
To find the limiting value as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{152}{1 + 4e^{-2x}} \][/tex]
- As [tex]\( x \)[/tex] approaches infinity, [tex]\( e^{-2x} \)[/tex] approaches 0.
- Thus, the denominator [tex]\( 1 + 4e^{-2x} \)[/tex] approaches 1 as [tex]\( x \)[/tex] increases indefinitely.
Therefore:
[tex]\[ \lim_{x \to \infty} f(x) = \frac{152}{1} = 152 \][/tex]
### Summary:
a. (Graph not shown here)
b. [tex]\( f(0) = 30.4 \)[/tex], [tex]\( f(10) \approx 152 \)[/tex]
c. The function is increasing.
d. The limiting value as [tex]\( x \)[/tex] approaches infinity is 152.
Regarding the graph options (A, B, C, D), choose the one that matches the description: an increasing curve starting at [tex]\( f(0) = 30.4 \)[/tex] and asymptotically approaching [tex]\( 152 \)[/tex] as [tex]\( x \)[/tex] approaches 10.
