Question

solve the rational inequality. write your answer in interval notation.

\\\frac{4x}{3-x} \ge 4x\\

Explanation:

Step1: Move all terms to one side

$$\frac{4x}{3-x} - 4x \geq 0$$

Step2: Find a common denominator

$$\frac{4x - 4x(3-x)}{3-x} \geq 0$$

Step3: Simplify the numerator

$$\frac{4x^2 - 8x}{3-x} \geq 0$$

Step4: Factor the numerator

$$\frac{4x(x-2)}{3-x} \geq 0$$

Step5: Identify critical points

$$x = 0, \quad x = 2, \quad x = 3$$

Step6: Test intervals for sign

For \(x \in (-\infty, 0)\), sign is negative.
For \(x \in (0, 2)\), sign is positive.
For \(x \in (2, 3)\), sign is negative.
For \(x \in (3, \infty)\), sign is positive.

Step7: Write the solution in interval notation

$$[0, 2] \cup (3, \infty)$$

Answer:

\([0, 2] \cup (3, \infty)\)