Answer :
To determine the conditions on [tex]\( a \)[/tex] and [tex]\( b \)[/tex] for the end behavior of the function [tex]\( f(x) = a(x+2)(x-3)^b \)[/tex] to be positive for both very large negative values of [tex]\( x \)[/tex] and very large positive values of [tex]\( x \)[/tex], we need to analyze the behavior of the function as [tex]\( x \to -\infty \)[/tex] and [tex]\( x \to \infty \)[/tex].
### 1. Behavior as [tex]\( x \to \infty \)[/tex]:
- When [tex]\( x \)[/tex] is very large and positive ([tex]\( x \gg 3 \)[/tex]):
- [tex]\( x + 2 \)[/tex] is positive.
- [tex]\( x - 3 \)[/tex] is positive.
- Since [tex]\( b \)[/tex] is positive, [tex]\( (x - 3)^b \)[/tex] is positive.
- The product [tex]\((x + 2)(x - 3)^b \)[/tex] will be positive.
Thus, for very large positive [tex]\( x \)[/tex], [tex]\( f(x) = a(x+2)(x-3)^b \)[/tex] will have the same sign as [tex]\( a \)[/tex] since both [tex]\( (x+2) \)[/tex] and [tex]\( (x-3)^b \)[/tex] are positive.
### 2. Behavior as [tex]\( x \to -\infty \)[/tex]:
- When [tex]\( x \)[/tex] is very large and negative ([tex]\( x \to -\infty \)[/tex]):
- [tex]\( x + 2 \)[/tex] is negative.
- [tex]\( x - 3 \)[/tex] is negative.
- The sign of [tex]\( (x-3)^b \)[/tex] depends on whether [tex]\( b \)[/tex] is even or odd:
- If [tex]\( b \)[/tex] is even, [tex]\((x-3)^b\)[/tex] will be positive.
- If [tex]\( b \)[/tex] is odd, [tex]\((x-3)^b\)[/tex] will be negative.
- If [tex]\( b \)[/tex] is even, [tex]\((x-3)^b\)[/tex] is positive, then the product [tex]\((x + 2)(x - 3)^b\)[/tex] will be negative [tex]\(\times\)[/tex] positive = negative.
- If [tex]\( b \)[/tex] is odd, [tex]\((x-3)^b\)[/tex] is negative, then the product [tex]\((x + 2)(x - 3)^b\)[/tex] will be negative [tex]\(\times\)[/tex] negative = positive.
### 3. Ensuring Positive End Behavior:
To ensure that [tex]\( f(x) \)[/tex] is positive at both ends:
- For very large positive values of [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] is positive if [tex]\( a \)[/tex] is positive.
- For very large negative values of [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] will also be positive only if the exponent [tex]\( b \)[/tex] is even which makes [tex]\( (x-3)^b \)[/tex] positive.
Thus, for the end behavior to be positive in both directions:
1. [tex]\( a \)[/tex] must be positive to ensure that the function doesn’t flip signs as [tex]\( x \to \infty \)[/tex].
2. [tex]\( b \)[/tex] must be even to ensure [tex]\( (x-3)^b \)[/tex] is positive when [tex]\( x \to -\infty \)[/tex].
Therefore, the correct statement is:
B. [tex]\( a \)[/tex] is positive, and [tex]\( b \)[/tex] is even.
### 1. Behavior as [tex]\( x \to \infty \)[/tex]:
- When [tex]\( x \)[/tex] is very large and positive ([tex]\( x \gg 3 \)[/tex]):
- [tex]\( x + 2 \)[/tex] is positive.
- [tex]\( x - 3 \)[/tex] is positive.
- Since [tex]\( b \)[/tex] is positive, [tex]\( (x - 3)^b \)[/tex] is positive.
- The product [tex]\((x + 2)(x - 3)^b \)[/tex] will be positive.
Thus, for very large positive [tex]\( x \)[/tex], [tex]\( f(x) = a(x+2)(x-3)^b \)[/tex] will have the same sign as [tex]\( a \)[/tex] since both [tex]\( (x+2) \)[/tex] and [tex]\( (x-3)^b \)[/tex] are positive.
### 2. Behavior as [tex]\( x \to -\infty \)[/tex]:
- When [tex]\( x \)[/tex] is very large and negative ([tex]\( x \to -\infty \)[/tex]):
- [tex]\( x + 2 \)[/tex] is negative.
- [tex]\( x - 3 \)[/tex] is negative.
- The sign of [tex]\( (x-3)^b \)[/tex] depends on whether [tex]\( b \)[/tex] is even or odd:
- If [tex]\( b \)[/tex] is even, [tex]\((x-3)^b\)[/tex] will be positive.
- If [tex]\( b \)[/tex] is odd, [tex]\((x-3)^b\)[/tex] will be negative.
- If [tex]\( b \)[/tex] is even, [tex]\((x-3)^b\)[/tex] is positive, then the product [tex]\((x + 2)(x - 3)^b\)[/tex] will be negative [tex]\(\times\)[/tex] positive = negative.
- If [tex]\( b \)[/tex] is odd, [tex]\((x-3)^b\)[/tex] is negative, then the product [tex]\((x + 2)(x - 3)^b\)[/tex] will be negative [tex]\(\times\)[/tex] negative = positive.
### 3. Ensuring Positive End Behavior:
To ensure that [tex]\( f(x) \)[/tex] is positive at both ends:
- For very large positive values of [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] is positive if [tex]\( a \)[/tex] is positive.
- For very large negative values of [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] will also be positive only if the exponent [tex]\( b \)[/tex] is even which makes [tex]\( (x-3)^b \)[/tex] positive.
Thus, for the end behavior to be positive in both directions:
1. [tex]\( a \)[/tex] must be positive to ensure that the function doesn’t flip signs as [tex]\( x \to \infty \)[/tex].
2. [tex]\( b \)[/tex] must be even to ensure [tex]\( (x-3)^b \)[/tex] is positive when [tex]\( x \to -\infty \)[/tex].
Therefore, the correct statement is:
B. [tex]\( a \)[/tex] is positive, and [tex]\( b \)[/tex] is even.
