Answer :
Certainly! Let's examine the function [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex] and analyze it thoroughly.
### Step 1: Define the Function
We are given the function:
[tex]\[ f(x) = \frac{3x^2 - 5}{x + 1} \][/tex]
### Step 2: Simplifying the Function
First, let's understand the components of the function:
- The numerator is [tex]\( 3x^2 - 5 \)[/tex].
- The denominator is [tex]\( x + 1 \)[/tex].
### Step 3: Identifying Domain
For the function [tex]\( f(x) \)[/tex] to be defined, the denominator must not be zero.
Hence, set [tex]\( x+1 \neq 0 \)[/tex]:
[tex]\[ x \neq -1 \][/tex]
So, the domain of the function is all real numbers except [tex]\( x = -1 \)[/tex]:
[tex]\[ x \in \mathbb{R} \setminus \{-1\} \][/tex]
### Step 4: Evaluating the Function
Let's look at evaluating the function at some key points to understand its behavior better.
1. At [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{3(0)^2 - 5}{0 + 1} = \frac{-5}{1} = -5 \][/tex]
2. At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{3(1)^2 - 5}{1 + 1} = \frac{3 - 5}{2} = \frac{-2}{2} = -1 \][/tex]
3. At [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \frac{3(2)^2 - 5}{2 + 1} = \frac{3 \cdot 4 - 5}{3} = \frac{12 - 5}{3} = \frac{7}{3} \approx 2.33 \][/tex]
### Step 5: Behavior Around [tex]\( x = -1 \)[/tex]
As [tex]\( x \)[/tex] approaches -1, the denominator approaches 0, which causes the function to exhibit a vertical asymptote. To see this more clearly:
- As [tex]\( x \)[/tex] approaches -1 from the right ([tex]\( x \to -1^+ \)[/tex]):
The value of [tex]\( f(x) \)[/tex] tends towards [tex]\( +\infty \)[/tex] since the fraction becomes increasingly positive.
- As [tex]\( x \)[/tex] approaches -1 from the left ([tex]\( x \to -1^- \)[/tex]):
The value of [tex]\( f(x) \)[/tex] tends towards [tex]\( -\infty \)[/tex] as the fraction becomes increasingly negative.
### Step 6: Analyzing the End Behavior
To analyze the end behavior, observe the function as [tex]\( x \)[/tex] becomes very large (positive or negative):
[tex]\[ \lim_{{x \to \infty}} f(x) \quad \text{and} \quad \lim_{{x \to -\infty}} f(x) \][/tex]
For large values of [tex]\( x \)[/tex]:
[tex]\[ f(x) \approx \frac{3x^2}{x} = 3x \][/tex]
Thus:
[tex]\[ \lim_{{x \to \infty}} f(x) = \infty \][/tex]
[tex]\[ \lim_{{x \to -\infty}} f(x) = -\infty \][/tex]
### Step 7: Summary
- Function: [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex]
- Domain: [tex]\( x \in \mathbb{R} \setminus \{-1\} \)[/tex]
- Key Points: [tex]\( f(0) = -5 \)[/tex], [tex]\( f(1) = -1 \)[/tex], [tex]\( f(2) \approx 2.33 \)[/tex]
- Vertical Asymptote: [tex]\( x = -1 \)[/tex]
- End Behavior: As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex]; as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]
This detailed analysis provides you with a comprehensive understanding of the behavior and characteristics of the function [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex].
### Step 1: Define the Function
We are given the function:
[tex]\[ f(x) = \frac{3x^2 - 5}{x + 1} \][/tex]
### Step 2: Simplifying the Function
First, let's understand the components of the function:
- The numerator is [tex]\( 3x^2 - 5 \)[/tex].
- The denominator is [tex]\( x + 1 \)[/tex].
### Step 3: Identifying Domain
For the function [tex]\( f(x) \)[/tex] to be defined, the denominator must not be zero.
Hence, set [tex]\( x+1 \neq 0 \)[/tex]:
[tex]\[ x \neq -1 \][/tex]
So, the domain of the function is all real numbers except [tex]\( x = -1 \)[/tex]:
[tex]\[ x \in \mathbb{R} \setminus \{-1\} \][/tex]
### Step 4: Evaluating the Function
Let's look at evaluating the function at some key points to understand its behavior better.
1. At [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{3(0)^2 - 5}{0 + 1} = \frac{-5}{1} = -5 \][/tex]
2. At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{3(1)^2 - 5}{1 + 1} = \frac{3 - 5}{2} = \frac{-2}{2} = -1 \][/tex]
3. At [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \frac{3(2)^2 - 5}{2 + 1} = \frac{3 \cdot 4 - 5}{3} = \frac{12 - 5}{3} = \frac{7}{3} \approx 2.33 \][/tex]
### Step 5: Behavior Around [tex]\( x = -1 \)[/tex]
As [tex]\( x \)[/tex] approaches -1, the denominator approaches 0, which causes the function to exhibit a vertical asymptote. To see this more clearly:
- As [tex]\( x \)[/tex] approaches -1 from the right ([tex]\( x \to -1^+ \)[/tex]):
The value of [tex]\( f(x) \)[/tex] tends towards [tex]\( +\infty \)[/tex] since the fraction becomes increasingly positive.
- As [tex]\( x \)[/tex] approaches -1 from the left ([tex]\( x \to -1^- \)[/tex]):
The value of [tex]\( f(x) \)[/tex] tends towards [tex]\( -\infty \)[/tex] as the fraction becomes increasingly negative.
### Step 6: Analyzing the End Behavior
To analyze the end behavior, observe the function as [tex]\( x \)[/tex] becomes very large (positive or negative):
[tex]\[ \lim_{{x \to \infty}} f(x) \quad \text{and} \quad \lim_{{x \to -\infty}} f(x) \][/tex]
For large values of [tex]\( x \)[/tex]:
[tex]\[ f(x) \approx \frac{3x^2}{x} = 3x \][/tex]
Thus:
[tex]\[ \lim_{{x \to \infty}} f(x) = \infty \][/tex]
[tex]\[ \lim_{{x \to -\infty}} f(x) = -\infty \][/tex]
### Step 7: Summary
- Function: [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex]
- Domain: [tex]\( x \in \mathbb{R} \setminus \{-1\} \)[/tex]
- Key Points: [tex]\( f(0) = -5 \)[/tex], [tex]\( f(1) = -1 \)[/tex], [tex]\( f(2) \approx 2.33 \)[/tex]
- Vertical Asymptote: [tex]\( x = -1 \)[/tex]
- End Behavior: As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex]; as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]
This detailed analysis provides you with a comprehensive understanding of the behavior and characteristics of the function [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex].
