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To analyze the function [tex]\( f(x) = \frac{\sqrt{2}}{x + 1} \)[/tex], we will look at its key characteristics and behavior for various values of [tex]\( x \)[/tex].
### 1. Domain of the Function
The function [tex]\( f(x) = \frac{\sqrt{2}}{x + 1} \)[/tex] is defined for all [tex]\( x \)[/tex] except where the denominator equals zero. The denominator [tex]\( x + 1 \)[/tex] equals zero when [tex]\( x = -1 \)[/tex]. Therefore, the domain of [tex]\( f(x) \)[/tex] is:
[tex]\[ \text{Domain} = \{ x \in \mathbb{R} \, | \, x \neq -1 \} \][/tex]
### 2. Evaluation of the Function at Specific Points
Let's compute [tex]\( f(x) \)[/tex] for some specific values of [tex]\( x \)[/tex]:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = \frac{\sqrt{2}}{-2 + 1} = \frac{\sqrt{2}}{-1} = -\sqrt{2} \][/tex]
- For [tex]\( x = -1 \)[/tex] (note: [tex]\( x = -1 \)[/tex] is not in the domain)
[tex]\[ f(-1) = \text{undefined} \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{\sqrt{2}}{0 + 1} = \frac{\sqrt{2}}{1} = \sqrt{2} \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{\sqrt{2}}{1 + 1} = \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \frac{\sqrt{2}}{2 + 1} = \frac{\sqrt{2}}{3} \][/tex]
### 3. Behavior and Plotting
It’s also helpful to understand the general behavior of the function:
- As [tex]\( x \)[/tex] approaches -1 from the left (i.e., [tex]\( x \rightarrow -1^- \)[/tex]), [tex]\( f(x) \rightarrow -\infty \)[/tex].
- As [tex]\( x \)[/tex] approaches -1 from the right (i.e., [tex]\( x \rightarrow -1^+ \)[/tex]), [tex]\( f(x) \rightarrow \infty \)[/tex].
- As [tex]\( x \)[/tex] becomes very large positively (i.e., [tex]\( x \rightarrow \infty \)[/tex]), [tex]\( f(x) \rightarrow 0^+ \)[/tex].
- As [tex]\( x \)[/tex] becomes very large negatively (i.e., [tex]\( x \rightarrow -\infty \)[/tex]), [tex]\( f(x) \rightarrow 0^- \)[/tex].
### 4. Summary
In conclusion, the function [tex]\( f(x) = \frac{\sqrt{2}}{x + 1} \)[/tex] is defined for all real numbers except [tex]\( x = -1 \)[/tex]. It exhibits a vertical asymptote at [tex]\( x = -1 \)[/tex] and approaches 0 as [tex]\( x \)[/tex] moves towards positive or negative infinity.
This analysis shows us the behavior and defines the significant points on the graph of the function.
### 1. Domain of the Function
The function [tex]\( f(x) = \frac{\sqrt{2}}{x + 1} \)[/tex] is defined for all [tex]\( x \)[/tex] except where the denominator equals zero. The denominator [tex]\( x + 1 \)[/tex] equals zero when [tex]\( x = -1 \)[/tex]. Therefore, the domain of [tex]\( f(x) \)[/tex] is:
[tex]\[ \text{Domain} = \{ x \in \mathbb{R} \, | \, x \neq -1 \} \][/tex]
### 2. Evaluation of the Function at Specific Points
Let's compute [tex]\( f(x) \)[/tex] for some specific values of [tex]\( x \)[/tex]:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = \frac{\sqrt{2}}{-2 + 1} = \frac{\sqrt{2}}{-1} = -\sqrt{2} \][/tex]
- For [tex]\( x = -1 \)[/tex] (note: [tex]\( x = -1 \)[/tex] is not in the domain)
[tex]\[ f(-1) = \text{undefined} \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{\sqrt{2}}{0 + 1} = \frac{\sqrt{2}}{1} = \sqrt{2} \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{\sqrt{2}}{1 + 1} = \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \frac{\sqrt{2}}{2 + 1} = \frac{\sqrt{2}}{3} \][/tex]
### 3. Behavior and Plotting
It’s also helpful to understand the general behavior of the function:
- As [tex]\( x \)[/tex] approaches -1 from the left (i.e., [tex]\( x \rightarrow -1^- \)[/tex]), [tex]\( f(x) \rightarrow -\infty \)[/tex].
- As [tex]\( x \)[/tex] approaches -1 from the right (i.e., [tex]\( x \rightarrow -1^+ \)[/tex]), [tex]\( f(x) \rightarrow \infty \)[/tex].
- As [tex]\( x \)[/tex] becomes very large positively (i.e., [tex]\( x \rightarrow \infty \)[/tex]), [tex]\( f(x) \rightarrow 0^+ \)[/tex].
- As [tex]\( x \)[/tex] becomes very large negatively (i.e., [tex]\( x \rightarrow -\infty \)[/tex]), [tex]\( f(x) \rightarrow 0^- \)[/tex].
### 4. Summary
In conclusion, the function [tex]\( f(x) = \frac{\sqrt{2}}{x + 1} \)[/tex] is defined for all real numbers except [tex]\( x = -1 \)[/tex]. It exhibits a vertical asymptote at [tex]\( x = -1 \)[/tex] and approaches 0 as [tex]\( x \)[/tex] moves towards positive or negative infinity.
This analysis shows us the behavior and defines the significant points on the graph of the function.
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