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What is the range of the function [tex]f(x)=\frac{3x-5}{x+2}[/tex]?

A. [tex](-\infty,-2) \cup (-2, \infty)[/tex]
B. [tex](-\infty, 0) \cup (0, \infty)[/tex]
C. [tex]\left(-\infty, \frac{5}{3}\right) \cup \left(\frac{5}{3}, \infty\right)[/tex]
D. [tex](-\infty, 3) \cup (3, \infty)[/tex]
E. [tex](-\infty, \infty)[/tex]

Sagot :

GinnyAnsweridn
To find the range of the function [tex]\( f(x) = \frac{3x - 5}{x + 2} \)[/tex], we need to analyze its behavior and understand how the function values span across the real numbers. Let's walk through this step-by-step:

1. Identify any asymptotes:
- Vertical asymptote: The function [tex]\( f(x) \)[/tex] has a vertical asymptote where the denominator is zero, i.e., [tex]\( x + 2 = 0 \)[/tex]. Solving for [tex]\( x \)[/tex] gives us [tex]\( x = -2 \)[/tex].
- Horizontal asymptote: For large absolute values of [tex]\( x \)[/tex] (both positive and negative infinity), the degree of the numerator and the denominator are the same. The horizontal asymptote will be the ratio of the leading coefficients: [tex]\( \frac{3}{1} = 3 \)[/tex].

2. Critical Points and Behavior Analysis:
- The critical points are found by setting the derivative of [tex]\( f(x) \)[/tex] to zero, which yields no real values in our context.
- Evaluate the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches the vertical asymptote from the left and right:
- As [tex]\( x \to -2^- \)[/tex], [tex]\( f(x) \to +\infty \)[/tex]
- As [tex]\( x \to -2^+ \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]

3. Evaluate limits:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \approx 3 \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \approx 3 \)[/tex]

4. Range Analysis:
- For very large positive or negative [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] approaches [tex]\( 3 \)[/tex] but never equals [tex]\( 3 \)[/tex]. This indicates that [tex]\( 3 \)[/tex] is not included in the range.
- From the behavior around the vertical asymptote, the function can take on very large positive and negative values.

5. Conclusion:
- The range of [tex]\( f(x) \)[/tex] excludes [tex]\( 3 \)[/tex] because the function approaches but never reaches this value. Thus, [tex]\( f(x) \)[/tex] takes all real values except [tex]\( 3 \)[/tex].

Given these details, we can determine that the range of the function [tex]\( f(x) = \frac{3x - 5}{x + 2} \)[/tex] is:

[tex]\[ (-\infty, 3) \cup (3, \infty) \][/tex]

This matches option D. Therefore, the correct answer is:

[tex]\[ \boxed{D. (-\infty, 3) \cup (3, \infty)} \][/tex]
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