IDNLearn.com: Your trusted platform for finding reliable answers. Our community provides accurate and timely answers to help you understand and solve any issue.
Sagot :
To find the horizontal asymptote of the function [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex], we need to analyze its behavior as [tex]\( x \)[/tex] approaches infinity ([tex]\( x \to \infty \)[/tex]).
### Step-by-Step Solution:
1. Identify the function:
[tex]\[ f(x) = \frac{3x^2 - 5}{x + 1} \][/tex]
2. Understand the concept of horizontal asymptotes:
Horizontal asymptotes occur when [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] or [tex]\(-\infty\)[/tex] and provide a value that the function approaches. We are particularly interested in the behavior as [tex]\( x \to \infty \)[/tex].
3. Consider the behavior as [tex]\( x \to \infty \)[/tex]:
To find the horizontal asymptote, we take the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{3x^2 - 5}{x + 1} \][/tex]
4. Simplify the expression within the limit:
When [tex]\( x \to \infty \)[/tex], the highest degree terms in the numerator and denominator dominate the behavior of the function:
[tex]\[ \lim_{x \to \infty} \frac{3x^2 - 5}{x + 1} = \lim_{x \to \infty} \frac{3x^2}{x + 1} \approx \lim_{x \to \infty} \frac{3x^2}{x} \quad (\text{for large } x) \][/tex]
5. Divide numerator and denominator by [tex]\( x \)[/tex]:
Simplify by dividing the numerator and the denominator by [tex]\( x \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{3x^2 / x}{(x + 1) / x} = \lim_{x \to \infty} \frac{3x}{1 + \frac{1}{x}} \][/tex]
6. Evaluate the simplified limit:
As [tex]\( x \to \infty \)[/tex], the term [tex]\(\frac{1}{x}\)[/tex] approaches 0:
[tex]\[ \lim_{x \to \infty} \frac{3x}{1 + \frac{1}{x}} = \lim_{x \to \infty} \frac{3x}{1 + 0} = \lim_{x \to \infty} 3x \][/tex]
Since [tex]\( 3x \)[/tex] grows without bound as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} 3x = \infty \][/tex]
Thus, the function [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex] does not approach a finite value but instead increases without bound as [tex]\( x \to \infty \)[/tex].
### Conclusion:
The horizontal asymptote is:
[tex]\[ \lim_{x \to \infty} f(x) = \infty \][/tex]
Therefore, [tex]\( f(x) \to \infty \)[/tex] as [tex]\( x \to \infty \)[/tex], indicating that there is no finite horizontal asymptote but rather the function increases indefinitely.
### Step-by-Step Solution:
1. Identify the function:
[tex]\[ f(x) = \frac{3x^2 - 5}{x + 1} \][/tex]
2. Understand the concept of horizontal asymptotes:
Horizontal asymptotes occur when [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] or [tex]\(-\infty\)[/tex] and provide a value that the function approaches. We are particularly interested in the behavior as [tex]\( x \to \infty \)[/tex].
3. Consider the behavior as [tex]\( x \to \infty \)[/tex]:
To find the horizontal asymptote, we take the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{3x^2 - 5}{x + 1} \][/tex]
4. Simplify the expression within the limit:
When [tex]\( x \to \infty \)[/tex], the highest degree terms in the numerator and denominator dominate the behavior of the function:
[tex]\[ \lim_{x \to \infty} \frac{3x^2 - 5}{x + 1} = \lim_{x \to \infty} \frac{3x^2}{x + 1} \approx \lim_{x \to \infty} \frac{3x^2}{x} \quad (\text{for large } x) \][/tex]
5. Divide numerator and denominator by [tex]\( x \)[/tex]:
Simplify by dividing the numerator and the denominator by [tex]\( x \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{3x^2 / x}{(x + 1) / x} = \lim_{x \to \infty} \frac{3x}{1 + \frac{1}{x}} \][/tex]
6. Evaluate the simplified limit:
As [tex]\( x \to \infty \)[/tex], the term [tex]\(\frac{1}{x}\)[/tex] approaches 0:
[tex]\[ \lim_{x \to \infty} \frac{3x}{1 + \frac{1}{x}} = \lim_{x \to \infty} \frac{3x}{1 + 0} = \lim_{x \to \infty} 3x \][/tex]
Since [tex]\( 3x \)[/tex] grows without bound as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} 3x = \infty \][/tex]
Thus, the function [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex] does not approach a finite value but instead increases without bound as [tex]\( x \to \infty \)[/tex].
### Conclusion:
The horizontal asymptote is:
[tex]\[ \lim_{x \to \infty} f(x) = \infty \][/tex]
Therefore, [tex]\( f(x) \to \infty \)[/tex] as [tex]\( x \to \infty \)[/tex], indicating that there is no finite horizontal asymptote but rather the function increases indefinitely.
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Discover the answers you need at IDNLearn.com. Thanks for visiting, and come back soon for more valuable insights.
