Question

The range of the real valued function \( f(x) = \frac{x^2 + x + 1}{x} \) is

• A. \((-\infty, 1) \cup (1, \infty)\)
• B. \((-\infty, -1] \cup [1, \infty)\)
• C. \((-\infty, -2] \cup [3, \infty)\)
• D. \((-\infty, -1] \cup [3, \infty)\)

Hint:

When given a rational function like \( \frac{x^2 + x + 1}{x} \), try simplifying by dividing or rewriting to identify the range. Use inequalities such as AM-GM on expressions like \( x + \frac{1}{x} \) to determine bounds.

Answer 1

The Correct Option is D

Given function: \[ f(x) = \frac{x^2 + x + 1}{x} = x + 1 + \frac{1}{x} \] Let \( y = f(x) = x + 1 + \frac{1}{x} \) Rewriting: \[ y = x + \frac{1}{x} + 1 \Rightarrow y - 1 = x + \frac{1}{x} \] Now, analyze the expression \( z = x + \frac{1}{x} \). For \( x>0 \), we know by AM-GM inequality: \[ x + \frac{1}{x} \geq 2 \Rightarrow y \geq 3 \] For \( x<0 \), let’s define \( x = -a \), \( a>0 \), then: \[ x + \frac{1}{x} = -a - \frac{1}{a} \leq -2 \Rightarrow y \leq -1 \] Therefore, the range of \( y \) is: \[ (-\infty, -1] \cup [3, \infty) \]