Great follow-up! Let's clarify how the interval ((-4, \frac{1}{3})) affects the sign of the expression:

For the inequality:

[ \frac{3x - 1}{x + 4} \leq 0 ]

The Role of Signs in Intervals

We split the number line based on the critical points (x = -4) (where the denominator is zero) and (x = \frac{1}{3}) (where the numerator is zero). This divides the number line into three intervals:

1. ( (-\infty, -4) )
2. ( (-4, \frac{1}{3}) )
3. ( (\frac{1}{3}, \infty) )

Each interval could either make the expression positive, negative, or zero. Let’s check how the sign of the expression changes across these intervals.

Testing the Sign in Each Interval

Interval 1: ( (-\infty, -4) )

• Choose a test value: (x = -5) [ \frac{3(-5) - 1}{-5 + 4} = \frac{-15 - 1}{-1} = \frac{-16}{-1} = 16 ] The result is positive. So, the expression is positive on ( (-\infty, -4) ).

Interval 2: ( (-4, \frac{1}{3}) )

• Choose a test value: (x = 0) [ \frac{3(0) - 1}{0 + 4} = \frac{-1}{4} = -\frac{1}{4} ] The result is negative. So, the expression is negative on ( (-4, \frac{1}{3}) ).

Interval 3: ( (\frac{1}{3}, \infty) )

• Choose a test value: (x = 1) [ \frac{3(1) - 1}{1 + 4} = \frac{2}{5} ] The result is positive. So, the expression is positive on ( (\frac{1}{3}, \infty) ).

Conclusion about Signs:

• Interval ( (-\infty, -4) ): Positive.
• Interval ( (-4, \frac{1}{3}) ): Negative.
• Interval ( (\frac{1}{3}, \infty) ): Positive.

The inequality (\frac{3x - 1}{x + 4} \leq 0) is satisfied when the expression is negative or zero. This happens in the interval ( (-4, \frac{1}{3}] ), where the sign is negative or exactly zero at (x = \frac{1}{3}).

Final Solution:

The solution to the inequality is: [ x \in (-4, \frac{1}{3}] ]

• The interval ( (-4, \frac{1}{3}) ) is where the expression is negative.
• At ( x = \frac{1}{3} ), the expression equals zero, which is allowed by the inequality (\leq 0).

Important Note on Signs:

• The critical points split the number line into intervals, and the sign of the expression changes depending on the values of (x) in those intervals.
• The critical points do not directly change the sign, but they determine the points where the sign might change, and you test each interval accordingly.

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Would you like any further clarification on how signs affect inequalities?

Here are 5 more questions to explore:

1. How do the signs change if the inequality is strict, such as (< 0)?
2. What if the denominator changes? How does that affect the intervals and signs?
3. Can you solve another inequality involving a rational expression and check the sign changes?
4. How do signs change when both numerator and denominator have higher-degree polynomials?
5. How would you determine the signs if both numerator and denominator were negative?

Tip: Always test the sign in each interval created by the critical points to determine where the inequality holds true!