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Sagot :
To determine the end behavior of the function [tex]\( g(x) = 4|x-2| - 3 \)[/tex], let's analyze the function as [tex]\( x \)[/tex] approaches negative and positive infinity.
1. As [tex]\( x \)[/tex] approaches negative infinity:
When [tex]\( x \)[/tex] is a very large negative number, the term [tex]\(|x-2|\)[/tex] will also be a large positive number since the absolute value of a large negative number shifted by 2 is still large. Therefore, [tex]\(|x-2| \approx |x|\)[/tex] and thus behaves like [tex]\( |x| \)[/tex]. Consequently,
[tex]\[ g(x) \approx 4|x| - 3. \][/tex]
Since [tex]\( |x| \)[/tex] equals [tex]\( -x \)[/tex] when [tex]\( x \)[/tex] is negative, we have:
[tex]\[ |x-2| \approx -x \][/tex]
Thus,
[tex]\[ g(x) \approx 4(-x) - 3 = -4x - 3. \][/tex]
As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( -4x - 3 \)[/tex] will decrease without bound, approaching negative infinity.
Therefore, as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches negative infinity.
2. As [tex]\( x \)[/tex] approaches positive infinity:
When [tex]\( x \)[/tex] is a very large positive number, the term [tex]\(|x-2|\)[/tex] will be approximately equal to [tex]\( x \)[/tex] itself, because the shift by 2 becomes negligible for very large values of [tex]\( x \)[/tex]. Thus,
[tex]\[ |x-2| \approx x. \][/tex]
Then,
[tex]\[ g(x) \approx 4x - 3. \][/tex]
As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( 4x - 3 \)[/tex] will increase without bound, approaching positive infinity.
Therefore, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
In summary, the correct selections for the function [tex]\( g(x) = 4|x-2| - 3 \)[/tex] are:
As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches negative infinity.
As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
1. As [tex]\( x \)[/tex] approaches negative infinity:
When [tex]\( x \)[/tex] is a very large negative number, the term [tex]\(|x-2|\)[/tex] will also be a large positive number since the absolute value of a large negative number shifted by 2 is still large. Therefore, [tex]\(|x-2| \approx |x|\)[/tex] and thus behaves like [tex]\( |x| \)[/tex]. Consequently,
[tex]\[ g(x) \approx 4|x| - 3. \][/tex]
Since [tex]\( |x| \)[/tex] equals [tex]\( -x \)[/tex] when [tex]\( x \)[/tex] is negative, we have:
[tex]\[ |x-2| \approx -x \][/tex]
Thus,
[tex]\[ g(x) \approx 4(-x) - 3 = -4x - 3. \][/tex]
As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( -4x - 3 \)[/tex] will decrease without bound, approaching negative infinity.
Therefore, as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches negative infinity.
2. As [tex]\( x \)[/tex] approaches positive infinity:
When [tex]\( x \)[/tex] is a very large positive number, the term [tex]\(|x-2|\)[/tex] will be approximately equal to [tex]\( x \)[/tex] itself, because the shift by 2 becomes negligible for very large values of [tex]\( x \)[/tex]. Thus,
[tex]\[ |x-2| \approx x. \][/tex]
Then,
[tex]\[ g(x) \approx 4x - 3. \][/tex]
As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( 4x - 3 \)[/tex] will increase without bound, approaching positive infinity.
Therefore, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
In summary, the correct selections for the function [tex]\( g(x) = 4|x-2| - 3 \)[/tex] are:
As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches negative infinity.
As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
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