Math Problem Statement

Consider the function � ( � )

� + 3 � − 1 f(x)= x−1 x+3 ​ . It holds � ( � ) ≥ 0 f(x)≥0 in:

To find where the function � ( � ) ≥ 0 f(x)≥0, we need to determine where the numerator and denominator have the same sign. � ( � )

� + 3 � − 1 f(x)= x−1 x+3 ​

� + 3 ≥ 0 x+3≥0 implies � ≥ − 3 x≥−3 � − 1

0 x−1>0 implies �

1 x>1 � − 1 < 0 x−1<0 implies � < 1 x<1 Combining these:

� + 3 � − 1 ≥ 0 x−1 x+3 ​ ≥0 in ( − ∞ , − 3 ] ∪ [ 1 , ∞ ) (−∞,−3]∪[1,∞) Therefore, the correct answer is ( − ∞ , − 3 ] ∪ [ 1 , ∞ ) (−∞,−3]∪[1,∞).

Solution

It looks like you're analyzing the rational function:

f(x)=x1x+3f(x) = \frac{x - 1}{x + 3}

and determining the intervals where f(x)0f(x) \geq 0.

The reasoning you've laid out involves finding the intervals where the numerator and the denominator have the same sign (both positive or both negative), since a fraction is non-negative when both parts are either both positive or both negative.

Steps:

  1. Numerator x1x - 1:

    • The numerator is zero at x=1x = 1.
    • It is positive for x>1x > 1, and negative for x<1x < 1.
  2. Denominator x+3x + 3:

    • The denominator is zero at x=3x = -3.
    • It is positive for x>3x > -3, and negative for x<3x < -3.

To find the solution, we need to examine the signs of the numerator and denominator in the intervals determined by the critical points x=3x = -3 and x=1x = 1:

  • Interval 1: (,3)(-\infty, -3)

    • The numerator x1x - 1 is negative, and the denominator x+3x + 3 is negative, so the fraction is positive.
  • Interval 2: (3,1)(-3, 1)

    • The numerator x1x - 1 is negative, and the denominator x+3x + 3 is positive, so the fraction is negative.
  • Interval 3: (1,)(1, \infty)

    • The numerator x1x - 1 is positive, and the denominator x+3x + 3 is positive, so the fraction is positive.

At x=3x = -3, the function is undefined due to division by zero, so we exclude x=3x = -3 from the solution. At x=1x = 1, the function is f(1)=0f(1) = 0, so x=1x = 1 is included in the solution.

Final Solution:

The solution is where f(x)0f(x) \geq 0, which is:

(,3)[1,)(-\infty, -3) \cup [1, \infty)

Questions for further exploration:

  1. What is the behavior of the function as xx approaches the asymptotes at x=3x = -3?
  2. How would the solution change if the inequality was f(x)>0f(x) > 0 instead of f(x)0f(x) \geq 0?
  3. Can you find the vertical asymptote and horizontal asymptote of the function f(x)f(x)?
  4. How would the graph of this function look for the intervals (,3)(-\infty, -3) and [1,)[1, \infty)?
  5. What would happen to the function if you added or subtracted a constant to the numerator or denominator?

Tip:

When solving inequalities involving rational functions, always carefully consider where the function is undefined (denominator equals zero), and check intervals between critical points where the numerator and denominator change sign.

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Inequalities
Sign Analysis

Formulas

f(x) = (x - 1) / (x + 3)

Theorems

Sign analysis of rational functions
Critical points and intervals of a rational function

Suitable Grade Level

Grades 9-11