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A population's instantaneous growth rate is the rate at which it grows for every instant in time. Function [tex]$r$[/tex] gives the instantaneous growth rate of a fruit fly population [tex]$x$[/tex] days after the start of an experiment.

[tex]\[ r(x) = 0.05\left(x^2+1\right)(x-6) \][/tex]

Consider the graph of function [tex]$r$[/tex].

Use the graph to complete each statement:

- Function [tex]$r$[/tex] has [tex]$\square$[/tex] and [tex]$\square$[/tex] two real zeros.
- Based on the instantaneous growth rate, the population decreased between 0 and [tex]$6 \checkmark$[/tex] days and the population increased after [tex]$6 \quad \sim$[/tex] days.



Answer :

To analyze the function [tex]\( r(x) = 0.05(x^2 + 1)(x - 6) \)[/tex] and determine the behavior of the population based on its instantaneous growth rate, we need to identify the zeros of the function and understand the intervals where the function is positive or negative.

### Step-by-Step Solution

1. Identify Potential Zeros:
The function [tex]\( r(x) = 0.05(x^2 + 1)(x - 6) \)[/tex] is determined by solving [tex]\( r(x) = 0 \)[/tex].
- There are two factors:
[tex]\[ 0.05(x^2 + 1) \quad \text{and} \quad (x - 6) \][/tex]
- Setting each factor to zero gives us the potential zeros:
[tex]\[ x^2 + 1 = 0 \quad \text{and} \quad x - 6 = 0 \][/tex]
- Solving [tex]\( x^2 + 1 = 0 \)[/tex] we get:
[tex]\[ x^2 = -1 \Rightarrow \text{No real solutions (as the square of a real number cannot be negative)} \][/tex]
- Solving [tex]\( x - 6 = 0 \)[/tex] we get:
[tex]\[ x = 6 \][/tex]

2. Determine Real Zeros:
- Thus, the only real zero from [tex]\( r(x) \)[/tex] is [tex]\( x = 6 \)[/tex].

3. Analyze the Interval Signs:
- To determine whether [tex]\( r(x) \)[/tex] is positive or negative around the zero, look at intervals defined by the zero [tex]\( x = 6 \)[/tex]:
[tex]\[ (x < 6) \quad \text{and} \quad (x > 6) \][/tex]

4. Local Behavior of The Function (Instantaneous Growth Rate):
- For [tex]\( x < 6 \)[/tex]:
- The term [tex]\( x - 6 \)[/tex] is negative.
- The term [tex]\( x^2 + 1 \)[/tex] is always positive (since any real number squared plus 1 is positive).
- Combining these: [tex]\( 0.05(x^2 + 1)(x - 6) \)[/tex] is negative.
- Therefore, [tex]\( r(x) < 0 \)[/tex] for [tex]\( x < 6 \)[/tex], indicating that the population decreases.
- For [tex]\( x > 6 \)[/tex]:
- The term [tex]\( x - 6 \)[/tex] is positive.
- Again, the term [tex]\( x^2 + 1 \)[/tex] is positive.
- Combining these: [tex]\( 0.05(x^2 + 1)(x - 6) \)[/tex] is positive.
- Therefore, [tex]\( r(x) > 0 \)[/tex] for [tex]\( x > 6 \)[/tex], indicating that the population increases.

### Complete The Statements:
1. Zeros:
- Function [tex]\( r \)[/tex] has one real zero (since [tex]\( x = 6 \)[/tex] is the only real zero).

2. Population Behavior:
- Based on the instantaneous growth rate, the population decreased between 0 and [tex]\( \boxed{6} \)[/tex] days and the population increased after [tex]\( \boxed{6} \)[/tex] days.