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Sagot :
To determine which functions are negative over the interval [tex]\([ -1, 1] \)[/tex], let's evaluate each function within this interval and determine if it ever takes on negative values.
Given functions are:
1. [tex]\( f(x) = x \)[/tex]
2. [tex]\( f(x) = x^2 \)[/tex]
3. [tex]\( f(x) = -x \)[/tex]
4. [tex]\( f(x) = \sin(x) \)[/tex]
Let's analyze each function:
1. Function [tex]\( f(x) = x \)[/tex]:
- Over the interval [tex]\([ -1, 1] \)[/tex], [tex]\( x \)[/tex] takes all values from [tex]\(-1\)[/tex] to [tex]\(1\)[/tex].
- Thus, [tex]\( f(x) = x \)[/tex] will be negative when [tex]\( x \)[/tex] is negative, i.e., for [tex]\( x \in [-1, 0)\)[/tex]. Therefore, [tex]\( f(x) = x \)[/tex] is negative over part of the interval [tex]\([-1, 1] \)[/tex].
2. Function [tex]\( f(x) = x^2 \)[/tex]:
- Over the interval [tex]\([ -1, 1] \)[/tex], [tex]\( x^2 \)[/tex] is always non-negative since squaring any real number results in a non-negative value.
- In this case, [tex]\( x^2 \)[/tex] produces values from 0 to 1 across the interval.
- Therefore, [tex]\( f(x) = x^2 \)[/tex] is never negative in the interval [tex]\([-1, 1] \)[/tex].
3. Function [tex]\( f(x) = -x \)[/tex]:
- Over the interval [tex]\([ -1, 1] \)[/tex], the value of [tex]\( -x \)[/tex] will be negative when [tex]\( x \in (0, 1] \)[/tex].
- Therefore, [tex]\( f(x) = -x \)[/tex] is negative over part of the interval [tex]\([-1, 1] \)[/tex].
4. Function [tex]\( f(x) = \sin(x) \)[/tex]:
- Over the interval [tex]\([ -1, 1] \)[/tex], the sine function [tex]\(\sin(x)\)[/tex] takes values that oscillate between [tex]\(\sin(-1)\)[/tex] and [tex]\(\sin(1)\)[/tex].
- Since the sine function oscillates, it will be negative in some portions of this interval.
- Hence, [tex]\( f(x) = \sin(x) \)[/tex] is negative over part of the interval [tex]\([-1, 1] \)[/tex].
In conclusion, the functions [tex]\( f(x) = x \)[/tex], [tex]\( f(x) = -x \)[/tex], and [tex]\( f(x) = \sin(x) \)[/tex] are negative for some portion of the interval [tex]\([ -1, 1] \)[/tex], while [tex]\( f(x) = x^2 \)[/tex] is never negative within this interval.
Given functions are:
1. [tex]\( f(x) = x \)[/tex]
2. [tex]\( f(x) = x^2 \)[/tex]
3. [tex]\( f(x) = -x \)[/tex]
4. [tex]\( f(x) = \sin(x) \)[/tex]
Let's analyze each function:
1. Function [tex]\( f(x) = x \)[/tex]:
- Over the interval [tex]\([ -1, 1] \)[/tex], [tex]\( x \)[/tex] takes all values from [tex]\(-1\)[/tex] to [tex]\(1\)[/tex].
- Thus, [tex]\( f(x) = x \)[/tex] will be negative when [tex]\( x \)[/tex] is negative, i.e., for [tex]\( x \in [-1, 0)\)[/tex]. Therefore, [tex]\( f(x) = x \)[/tex] is negative over part of the interval [tex]\([-1, 1] \)[/tex].
2. Function [tex]\( f(x) = x^2 \)[/tex]:
- Over the interval [tex]\([ -1, 1] \)[/tex], [tex]\( x^2 \)[/tex] is always non-negative since squaring any real number results in a non-negative value.
- In this case, [tex]\( x^2 \)[/tex] produces values from 0 to 1 across the interval.
- Therefore, [tex]\( f(x) = x^2 \)[/tex] is never negative in the interval [tex]\([-1, 1] \)[/tex].
3. Function [tex]\( f(x) = -x \)[/tex]:
- Over the interval [tex]\([ -1, 1] \)[/tex], the value of [tex]\( -x \)[/tex] will be negative when [tex]\( x \in (0, 1] \)[/tex].
- Therefore, [tex]\( f(x) = -x \)[/tex] is negative over part of the interval [tex]\([-1, 1] \)[/tex].
4. Function [tex]\( f(x) = \sin(x) \)[/tex]:
- Over the interval [tex]\([ -1, 1] \)[/tex], the sine function [tex]\(\sin(x)\)[/tex] takes values that oscillate between [tex]\(\sin(-1)\)[/tex] and [tex]\(\sin(1)\)[/tex].
- Since the sine function oscillates, it will be negative in some portions of this interval.
- Hence, [tex]\( f(x) = \sin(x) \)[/tex] is negative over part of the interval [tex]\([-1, 1] \)[/tex].
In conclusion, the functions [tex]\( f(x) = x \)[/tex], [tex]\( f(x) = -x \)[/tex], and [tex]\( f(x) = \sin(x) \)[/tex] are negative for some portion of the interval [tex]\([ -1, 1] \)[/tex], while [tex]\( f(x) = x^2 \)[/tex] is never negative within this interval.
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