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Sagot :
To understand the end behavior of the function [tex]\( g(x) = 4|x-2| - 3 \)[/tex], let's analyze what happens as [tex]\( x \)[/tex] approaches negative infinity ([tex]\( -\infty \)[/tex]) and positive infinity ([tex]\( \infty \)[/tex]).
### Step-by-Step Analysis:
1. Consider the function as [tex]\( x \)[/tex] approaches negative infinity ([tex]\( -\infty \)[/tex]):
- The term [tex]\( |x-2| \)[/tex] represents the absolute value of [tex]\( x-2 \)[/tex].
- As [tex]\( x \)[/tex] becomes very large in the negative direction ([tex]\( x \to -\infty \)[/tex]), [tex]\( |x-2| \)[/tex] essentially behaves like [tex]\( |-x| \)[/tex] because [tex]\(|x-2| \)[/tex] accounts for the absolute distance from 2.
- Therefore, [tex]\( |x-2| \)[/tex] will tend towards positive infinity as [tex]\( x \to -\infty \)[/tex].
2. Evaluate the effect on the function [tex]\( g(x) \)[/tex]:
- As [tex]\( |x-2| \)[/tex] approaches positive infinity ([tex]\( \infty \)[/tex]), the term [tex]\( 4|x-2| \)[/tex] will also approach positive infinity because multiplying a large positive number by 4 results in an even larger positive number.
- Subtracting 3 from this very large number ([tex]\( 4|x-2| - 3 \)[/tex]) will still result in a very large positive number.
- Hence, as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex], [tex]\( g(x) \)[/tex] approaches positive infinity.
3. Repeat the analysis for [tex]\( x \)[/tex] approaching positive infinity ([tex]\( \infty \)[/tex]):
- As [tex]\( x \)[/tex] becomes very large in the positive direction ([tex]\( x \to \infty \)[/tex]), [tex]\( |x-2| \)[/tex] again behaves like [tex]\( x \)[/tex] since any large value of [tex]\( x \)[/tex] minus 2 will still yield a large positive value, and taking the absolute value will have no effect.
- So, in this case as well, [tex]\( |x-2| \)[/tex] will tend towards positive infinity.
4. Evaluate the effect on [tex]\( g(x) \)[/tex] once more:
- With [tex]\( |x-2| \to \infty \)[/tex], the term [tex]\( 4|x-2| \)[/tex] will again approach positive infinity.
- Subtracting 3 from this very large number ([tex]\( 4|x-2| - 3\)[/tex]) still results in a very large positive number.
- Hence, as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex], [tex]\( g(x) \)[/tex] also approaches positive infinity.
### Summary:
- As [tex]\( x \)[/tex] approaches negative infinity ([tex]\( -\infty \)[/tex]), [tex]\( g(x) \)[/tex] approaches positive infinity.
- As [tex]\( x \)[/tex] approaches positive infinity ([tex]\( \infty \)[/tex]), [tex]\( g(x) \)[/tex] also approaches positive infinity.
Therefore, the correct answers for each drop-down menu are:
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
### Step-by-Step Analysis:
1. Consider the function as [tex]\( x \)[/tex] approaches negative infinity ([tex]\( -\infty \)[/tex]):
- The term [tex]\( |x-2| \)[/tex] represents the absolute value of [tex]\( x-2 \)[/tex].
- As [tex]\( x \)[/tex] becomes very large in the negative direction ([tex]\( x \to -\infty \)[/tex]), [tex]\( |x-2| \)[/tex] essentially behaves like [tex]\( |-x| \)[/tex] because [tex]\(|x-2| \)[/tex] accounts for the absolute distance from 2.
- Therefore, [tex]\( |x-2| \)[/tex] will tend towards positive infinity as [tex]\( x \to -\infty \)[/tex].
2. Evaluate the effect on the function [tex]\( g(x) \)[/tex]:
- As [tex]\( |x-2| \)[/tex] approaches positive infinity ([tex]\( \infty \)[/tex]), the term [tex]\( 4|x-2| \)[/tex] will also approach positive infinity because multiplying a large positive number by 4 results in an even larger positive number.
- Subtracting 3 from this very large number ([tex]\( 4|x-2| - 3 \)[/tex]) will still result in a very large positive number.
- Hence, as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex], [tex]\( g(x) \)[/tex] approaches positive infinity.
3. Repeat the analysis for [tex]\( x \)[/tex] approaching positive infinity ([tex]\( \infty \)[/tex]):
- As [tex]\( x \)[/tex] becomes very large in the positive direction ([tex]\( x \to \infty \)[/tex]), [tex]\( |x-2| \)[/tex] again behaves like [tex]\( x \)[/tex] since any large value of [tex]\( x \)[/tex] minus 2 will still yield a large positive value, and taking the absolute value will have no effect.
- So, in this case as well, [tex]\( |x-2| \)[/tex] will tend towards positive infinity.
4. Evaluate the effect on [tex]\( g(x) \)[/tex] once more:
- With [tex]\( |x-2| \to \infty \)[/tex], the term [tex]\( 4|x-2| \)[/tex] will again approach positive infinity.
- Subtracting 3 from this very large number ([tex]\( 4|x-2| - 3\)[/tex]) still results in a very large positive number.
- Hence, as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex], [tex]\( g(x) \)[/tex] also approaches positive infinity.
### Summary:
- As [tex]\( x \)[/tex] approaches negative infinity ([tex]\( -\infty \)[/tex]), [tex]\( g(x) \)[/tex] approaches positive infinity.
- As [tex]\( x \)[/tex] approaches positive infinity ([tex]\( \infty \)[/tex]), [tex]\( g(x) \)[/tex] also approaches positive infinity.
Therefore, the correct answers for each drop-down menu are:
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
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