Answer :
Given the function [tex]\( r(x) = 0.05(x^2 + 1)(x - 6) \)[/tex], we can determine certain characteristics of the population growth over time by analyzing where the function intersects the x-axis (its zeros) and the behavior of the function at specific points.
### Zeros of the Function
#### Complex Zeros
Analyzing the function, [tex]\( r(x) \)[/tex]:
- [tex]\( r(x) = 0 \)[/tex] when [tex]\( x^2 + 1 = 0 \)[/tex] or [tex]\( x - 6 = 0 \)[/tex].
- The quadratic [tex]\( x^2 + 1 = 0 \implies x^2 = -1 \)[/tex], which gives solutions [tex]\( x = \pm i \)[/tex], where [tex]\( i \)[/tex] is the imaginary unit. These are complex roots.
#### Real Zeros
- [tex]\( x - 6 = 0 \implies x = 6 \)[/tex]. This is a real root.
- Thus, function [tex]\( r \)[/tex] has exactly one real zero.
### Growth Rate at [tex]\( x = 6 \)[/tex]
- To find whether the population increased or decreased after day 6, we substitute [tex]\( x = 6 \)[/tex] into the function.
- [tex]\( r(6) = 0.05(6^2 + 1)(6 - 6) = 0.05(37)(0) = 0 \)[/tex].
Since [tex]\( r(6) = 0 \)[/tex], the instantaneous growth rate at day 6 is zero, implying no change in population growth exactly at day 6. However, to decide if the population increased after day 6, we need to analyze values slightly greater than 6. Given that the coefficient of [tex]\( x^2 \)[/tex] is positive in the quadratic term [tex]\( x^2 + 1 \)[/tex], the overall function [tex]\( r(x) = 0.05(x^2 + 1)(x - 6) \)[/tex] will start negative for values slightly greater than 6, indicating a decrease in the population rate after day 6.
### Summary of Statements:
1. Function [tex]\( r \)[/tex] has two complex zeros.
2. Function [tex]\( r \)[/tex] has one real zero.
3. The population decreased after day 6.
In conclusion:
Function [tex]\( r \)[/tex] has two complex zeros, one real zero, and the population decreased after 6 days.
### Zeros of the Function
#### Complex Zeros
Analyzing the function, [tex]\( r(x) \)[/tex]:
- [tex]\( r(x) = 0 \)[/tex] when [tex]\( x^2 + 1 = 0 \)[/tex] or [tex]\( x - 6 = 0 \)[/tex].
- The quadratic [tex]\( x^2 + 1 = 0 \implies x^2 = -1 \)[/tex], which gives solutions [tex]\( x = \pm i \)[/tex], where [tex]\( i \)[/tex] is the imaginary unit. These are complex roots.
#### Real Zeros
- [tex]\( x - 6 = 0 \implies x = 6 \)[/tex]. This is a real root.
- Thus, function [tex]\( r \)[/tex] has exactly one real zero.
### Growth Rate at [tex]\( x = 6 \)[/tex]
- To find whether the population increased or decreased after day 6, we substitute [tex]\( x = 6 \)[/tex] into the function.
- [tex]\( r(6) = 0.05(6^2 + 1)(6 - 6) = 0.05(37)(0) = 0 \)[/tex].
Since [tex]\( r(6) = 0 \)[/tex], the instantaneous growth rate at day 6 is zero, implying no change in population growth exactly at day 6. However, to decide if the population increased after day 6, we need to analyze values slightly greater than 6. Given that the coefficient of [tex]\( x^2 \)[/tex] is positive in the quadratic term [tex]\( x^2 + 1 \)[/tex], the overall function [tex]\( r(x) = 0.05(x^2 + 1)(x - 6) \)[/tex] will start negative for values slightly greater than 6, indicating a decrease in the population rate after day 6.
### Summary of Statements:
1. Function [tex]\( r \)[/tex] has two complex zeros.
2. Function [tex]\( r \)[/tex] has one real zero.
3. The population decreased after day 6.
In conclusion:
Function [tex]\( r \)[/tex] has two complex zeros, one real zero, and the population decreased after 6 days.
