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Sagot :
Sure, let's analyze the ranges and functions carefully step-by-step. We need to determine the values in the range of [tex]\( f(x) = \sqrt{x+1} - 1 \)[/tex] that do not fall within the given value ranges of [tex]\( g(x) \)[/tex].
1. Determine the domain and range of [tex]\( f(x) \)[/tex]
- The function [tex]\( f(x) = \sqrt{x+1} - 1 \)[/tex] is defined for values of [tex]\( x \)[/tex] such that the argument of the square root is non-negative.
- Therefore, [tex]\( x + 1 \geq 0 \)[/tex] which implies [tex]\( x \geq -1 \)[/tex].
- To find the range of [tex]\( f(x) \)[/tex]:
- When [tex]\( x = -1 \)[/tex], [tex]\( f(x) = \sqrt{-1 + 1} - 1 = 0 - 1 = -1 \)[/tex]
- As [tex]\( x \)[/tex] increases from [tex]\(-1\)[/tex] to [tex]\(\infty\)[/tex], [tex]\( \sqrt{x+1} \)[/tex] will also increase from 0 to [tex]\(\infty\)[/tex], and therefore [tex]\( f(x) \)[/tex] will increase from [tex]\(-1\)[/tex] to [tex]\(\infty\)[/tex].
- Therefore, the range of [tex]\( f(x) \)[/tex] is [tex]\([-1, \infty) \)[/tex].
2. Compare the range of [tex]\( f(x) \)[/tex] with the given ranges and find values that do not intersect
Let's analyze the given ranges one by one:
- [tex]\( [-1, 1) \)[/tex]: This range starts from [tex]\(-1\)[/tex] inclusive and extends to [tex]\(1\)[/tex] but not including [tex]\(1\)[/tex]. The intersection with [tex]\([-1, \infty)\)[/tex] is [tex]\([-1, 1)\)[/tex]. So, this range is a valid subset of [tex]\([-1, \infty)\)[/tex].
- [tex]\( (-1, 1) \)[/tex]: This range starts just above [tex]\(-1\)[/tex] and extends to but does not include [tex]\(1\)[/tex]. The intersection with [tex]\([-1, \infty)\)[/tex] is [tex]\((-1, 1)\)[/tex]. So, this range is also a valid subset of [tex]\([-1, \infty)\)[/tex].
- [tex]\( (-\infty, 1] \)[/tex]: This range extends from negative infinity up to and including [tex]\(1\)[/tex]. The intersection with [tex]\([-1, \infty)\)[/tex] would be [tex]\([-1, 1]\)[/tex], so this also overlaps with some part of the range of [tex]\( f(x) \)[/tex].
- [tex]\( (-\infty, 1) \)[/tex]: This range extends from negative infinity up to but not including [tex]\(1\)[/tex]. The intersection with [tex]\([-1, \infty)\)[/tex] would be [tex]\([-1, 1)\)[/tex], which also overlaps with some part of the range of [tex]\( f(x) \)[/tex].
Thus, all given ranges overlap with the range of [tex]\( f(x) \)[/tex]. Hence, there are no values in the range of [tex]\( f(x) = \sqrt{x+1} - 1 \)[/tex] that are not in the graph of [tex]\( g(x) \)[/tex].
Therefore, the correct answer is:
[tex]\[ -1 \][/tex]
1. Determine the domain and range of [tex]\( f(x) \)[/tex]
- The function [tex]\( f(x) = \sqrt{x+1} - 1 \)[/tex] is defined for values of [tex]\( x \)[/tex] such that the argument of the square root is non-negative.
- Therefore, [tex]\( x + 1 \geq 0 \)[/tex] which implies [tex]\( x \geq -1 \)[/tex].
- To find the range of [tex]\( f(x) \)[/tex]:
- When [tex]\( x = -1 \)[/tex], [tex]\( f(x) = \sqrt{-1 + 1} - 1 = 0 - 1 = -1 \)[/tex]
- As [tex]\( x \)[/tex] increases from [tex]\(-1\)[/tex] to [tex]\(\infty\)[/tex], [tex]\( \sqrt{x+1} \)[/tex] will also increase from 0 to [tex]\(\infty\)[/tex], and therefore [tex]\( f(x) \)[/tex] will increase from [tex]\(-1\)[/tex] to [tex]\(\infty\)[/tex].
- Therefore, the range of [tex]\( f(x) \)[/tex] is [tex]\([-1, \infty) \)[/tex].
2. Compare the range of [tex]\( f(x) \)[/tex] with the given ranges and find values that do not intersect
Let's analyze the given ranges one by one:
- [tex]\( [-1, 1) \)[/tex]: This range starts from [tex]\(-1\)[/tex] inclusive and extends to [tex]\(1\)[/tex] but not including [tex]\(1\)[/tex]. The intersection with [tex]\([-1, \infty)\)[/tex] is [tex]\([-1, 1)\)[/tex]. So, this range is a valid subset of [tex]\([-1, \infty)\)[/tex].
- [tex]\( (-1, 1) \)[/tex]: This range starts just above [tex]\(-1\)[/tex] and extends to but does not include [tex]\(1\)[/tex]. The intersection with [tex]\([-1, \infty)\)[/tex] is [tex]\((-1, 1)\)[/tex]. So, this range is also a valid subset of [tex]\([-1, \infty)\)[/tex].
- [tex]\( (-\infty, 1] \)[/tex]: This range extends from negative infinity up to and including [tex]\(1\)[/tex]. The intersection with [tex]\([-1, \infty)\)[/tex] would be [tex]\([-1, 1]\)[/tex], so this also overlaps with some part of the range of [tex]\( f(x) \)[/tex].
- [tex]\( (-\infty, 1) \)[/tex]: This range extends from negative infinity up to but not including [tex]\(1\)[/tex]. The intersection with [tex]\([-1, \infty)\)[/tex] would be [tex]\([-1, 1)\)[/tex], which also overlaps with some part of the range of [tex]\( f(x) \)[/tex].
Thus, all given ranges overlap with the range of [tex]\( f(x) \)[/tex]. Hence, there are no values in the range of [tex]\( f(x) = \sqrt{x+1} - 1 \)[/tex] that are not in the graph of [tex]\( g(x) \)[/tex].
Therefore, the correct answer is:
[tex]\[ -1 \][/tex]
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