Question

True or False. The graph of f ( x ) = \frac { x } { x - 3 } \text { is above } the \(x\) -axis for \(x<0\) or \(x>3,\) so the solution set of the inequality \(\frac{x}{x-3} \geq 0\) is \(\\{x \mid x \leq 0\) or \(x \geq 3\\}\)

Short answer

False. The correct solution set is \( \{ x \mid x \leq 0 \text{ or } x > 3 \} \).

Step-by-step solution

Identify critical points

Locate the points where the function equals zero or is undefined. For the function \( f(x) = \frac{x}{x-3} \), set the numerator equal to zero: \( x = 0 \). The function is undefined where the denominator is zero: \( x - 3 = 0 \Rightarrow x = 3 \).

Determine the sign of the function

Test values in the intervals determined by the critical points to see where the function is positive or negative. The intervals are: \(( -\infty, 0 )\), \(( 0, 3 )\), and \(( 3, +\infty )\). Use test points such as \( x = -1 \), \( x = 1 \), and \( x = 4 \) respectively.

Analyze the sign of the test points

For \( x = -1 \), \( f(-1) = \frac{-1}{-4} = \frac{1}{4} \text{ which is positive}.\) For \( x = 1 \), \( f(1) = \frac{1}{1-3} = \frac{1}{-2} = -\frac{1}{2} \text{ which is negative}.\) For \( x = 4 \), \( f(4) = \frac{4}{1} = 4 \text{ which is positive}.\)

Interpret the intervals

The function \( f(x) = \frac{x}{x-3} \) is positive for \( x < 0 \) and \( x > 3 \), and negative for \( 0 < x < 3 \).

Include critical points in the solution

The inequality \( \frac{x}{x-3} \geq 0 \) includes the points where the function equals zero, which is at \( x = 0 \). It should exclude the point where the function is undefined, which is at \( x = 3 \).

Write the solution set

Thus, the solution set of the inequality \( \frac{x}{x-3} \geq 0 \) is \( \{ x \mid x \leq 0 \text{ or } x > 3 \} \), not \( x \geq 3 \).

Key concepts

Critical Points

The first step in solving inequalities involving rational functions is to find the critical points. These are the values of x where the function is either zero or undefined. For the function given by: ds \[ f(x) = \frac{x}{x-3} \], we need to determine where the numerator and denominator equal zero. When the numerator is zero, the whole fraction becomes zero. Hence, we set the numerator x = 0. Therefore, our first critical point is at x = 0. Similarly, the function is undefined wherever the denominator equals zero. Setting the denominator equal to zero gives: \[ x - 3 = 0 \rightarrow x = 3 \]. So, our second critical point is x = 3.

Interval Testing

After finding the critical points, the next step is to divide the number line into intervals based on these points. These intervals help us analyze where our function is positive, negative, or zero. For our function, the critical points x = 0 and x = 3 divide the number line into three intervals:

• Interval 1: \(( -\text{∞}, 0 )\)
• Interval 2: \(( 0, 3 )\)
• Interval 3: \(( 3, +\text{∞} )\)

.To determine the sign of the function within these intervals, we use test points; a value of x within each interval. For example:

• Test point for Interval 1: x = -1
• Test point for Interval 2: x = 1
• Test point for Interval 3: x = 4

.

Sign Analysis

Sign analysis helps us determine whether our function is positive or negative within each interval. Let's evaluate our function at each test point:

• For x = -1: \[ f(-1) = \frac{-1}{-4} = \frac{1}{4} \] which is positive.
• For x = 1: \[ f(1) = \frac{1}{1-3} = \frac{1}{-2} = -\frac{1}{2} \] which is negative.
• For x = 4: \[ f(4) = \frac{4}{1} = 4 \] which is positive.

With this analysis, we see that the function is:

• Positive in Interval 1 \(( -\text{∞}, 0 )\)
• Negative in Interval 2 \(( 0, 3 )\)
• Positive again in Interval 3 \(( 3, +\text{∞} )\)

.

Solution Sets

The final step is to form the solution set based on our sign analysis. We need the function to be greater than or equal to zero. This means it should be positive or zero. From our analysis, we know:

• The function is positive for \( x < 0 \) and \( x > 3 \)
• The function is zero at \( x = 0 \)
• The function is undefined at \( x = 3 \)

. Thus, we include 0 in our solution set but exclude 3. Therefore, the solution set for the inequality:\[ \frac{x}{x-3} \geq 0 \] is: \{ x \mid x \leq 0 \text{ or } \ x > 3 \}. This includes all x less than or equal to zero and all x greater than 3.