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Sagot :
Certainly! Let's analyze the function [tex]\( g(x) = -\frac{1}{x} - 1 \)[/tex] step by step to answer each part of the question.
### Part (a): Write down the range of [tex]\( f \)[/tex]
To determine the range of [tex]\( g(x) \)[/tex]:
- Consider the term [tex]\(-\frac{1}{x}\)[/tex]. As [tex]\( x \)[/tex] can be any real number except zero, [tex]\(-\frac{1}{x}\)[/tex] can take any real value except zero.
- When [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex], [tex]\(-\frac{1}{x} \)[/tex] approaches 0. Consequently, [tex]\( g(x) = -\frac{1}{x} - 1 \)[/tex] approaches [tex]\(-1\)[/tex].
- Therefore, [tex]\( g(x) \)[/tex] can take any real number except [tex]\(-1\)[/tex].
So, the range of [tex]\( g(x) \)[/tex] is:
[tex]\[ \text{All real numbers except } -1 \][/tex]
### Part (b): Write down the horizontal asymptote of [tex]\( f \)[/tex]
The horizontal asymptote is the value that [tex]\( g(x) \)[/tex] approaches as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex], the term [tex]\(-\frac{1}{x} \)[/tex] approaches 0.
- Therefore, [tex]\( g(x) = -\frac{1}{x} - 1 \)[/tex] approaches [tex]\(-1\)[/tex].
So, the horizontal asymptote of [tex]\( g(x) \)[/tex] is:
[tex]\[ -1 \][/tex]
### Part (c): Determine the value of [tex]\( x \)[/tex] for which [tex]\( g(x) < 0 \)[/tex]
We need to find for which values of [tex]\( x \)[/tex] the inequality [tex]\( g(x) < 0 \)[/tex] holds true.
- [tex]\( g(x) < 0 \)[/tex] translates to [tex]\(-\frac{1}{x} - 1 < 0 \)[/tex].
- Rearranging the inequality: [tex]\(-\frac{1}{x} < 1 \)[/tex].
- This simplifies to: [tex]\(\frac{1}{x} > -1 \)[/tex].
Given this inequality:
- If [tex]\( x \)[/tex] is positive, [tex]\(\frac{1}{x}\)[/tex] is positive and always greater than [tex]\(-1\)[/tex].
- So, all positive values of [tex]\( x \)[/tex] will satisfy the inequality.
Thus, the value of [tex]\( x \)[/tex] for which [tex]\( g(x) < 0 \)[/tex] is:
[tex]\[ x > 0 \][/tex]
### Part (d): Write down the equation of [tex]\( h \)[/tex], where [tex]\( h(x) = -g(x) \)[/tex]
Given [tex]\( h(x) = -g(x) \)[/tex] and [tex]\( g(x) = -\frac{1}{x} - 1 \)[/tex], we perform the following calculation:
- [tex]\( g(x) = -\frac{1}{x} - 1 \)[/tex].
- Therefore, [tex]\( -g(x) = -\left(-\frac{1}{x} - 1\right) = \frac{1}{x} + 1 \)[/tex].
So, the equation of [tex]\( h(x) \)[/tex] is:
[tex]\[ h(x) = \frac{1}{x} + 1 \][/tex]
### Part (a): Write down the range of [tex]\( f \)[/tex]
To determine the range of [tex]\( g(x) \)[/tex]:
- Consider the term [tex]\(-\frac{1}{x}\)[/tex]. As [tex]\( x \)[/tex] can be any real number except zero, [tex]\(-\frac{1}{x}\)[/tex] can take any real value except zero.
- When [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex], [tex]\(-\frac{1}{x} \)[/tex] approaches 0. Consequently, [tex]\( g(x) = -\frac{1}{x} - 1 \)[/tex] approaches [tex]\(-1\)[/tex].
- Therefore, [tex]\( g(x) \)[/tex] can take any real number except [tex]\(-1\)[/tex].
So, the range of [tex]\( g(x) \)[/tex] is:
[tex]\[ \text{All real numbers except } -1 \][/tex]
### Part (b): Write down the horizontal asymptote of [tex]\( f \)[/tex]
The horizontal asymptote is the value that [tex]\( g(x) \)[/tex] approaches as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex], the term [tex]\(-\frac{1}{x} \)[/tex] approaches 0.
- Therefore, [tex]\( g(x) = -\frac{1}{x} - 1 \)[/tex] approaches [tex]\(-1\)[/tex].
So, the horizontal asymptote of [tex]\( g(x) \)[/tex] is:
[tex]\[ -1 \][/tex]
### Part (c): Determine the value of [tex]\( x \)[/tex] for which [tex]\( g(x) < 0 \)[/tex]
We need to find for which values of [tex]\( x \)[/tex] the inequality [tex]\( g(x) < 0 \)[/tex] holds true.
- [tex]\( g(x) < 0 \)[/tex] translates to [tex]\(-\frac{1}{x} - 1 < 0 \)[/tex].
- Rearranging the inequality: [tex]\(-\frac{1}{x} < 1 \)[/tex].
- This simplifies to: [tex]\(\frac{1}{x} > -1 \)[/tex].
Given this inequality:
- If [tex]\( x \)[/tex] is positive, [tex]\(\frac{1}{x}\)[/tex] is positive and always greater than [tex]\(-1\)[/tex].
- So, all positive values of [tex]\( x \)[/tex] will satisfy the inequality.
Thus, the value of [tex]\( x \)[/tex] for which [tex]\( g(x) < 0 \)[/tex] is:
[tex]\[ x > 0 \][/tex]
### Part (d): Write down the equation of [tex]\( h \)[/tex], where [tex]\( h(x) = -g(x) \)[/tex]
Given [tex]\( h(x) = -g(x) \)[/tex] and [tex]\( g(x) = -\frac{1}{x} - 1 \)[/tex], we perform the following calculation:
- [tex]\( g(x) = -\frac{1}{x} - 1 \)[/tex].
- Therefore, [tex]\( -g(x) = -\left(-\frac{1}{x} - 1\right) = \frac{1}{x} + 1 \)[/tex].
So, the equation of [tex]\( h(x) \)[/tex] is:
[tex]\[ h(x) = \frac{1}{x} + 1 \][/tex]
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