Question

solve the following rational inequality.\\(\frac{x^3 - 4x}{x^2 + 2} \leq 0\\)\\(-2 \leq x \leq 2\\)\\(x \leq -2\\) or \\(x \geq 2\\)\\(-2 \leq x \leq 0\\) or \\(x \geq 2\\)\\(x \leq -2\\) or \\(0 \leq x \leq 2\\)

Explanation:

Step1: Analyze the denominator

The denominator is \(x^2 + 2\). Since \(x^2\geq0\) for all real \(x\), then \(x^2 + 2\geq2>0\) for all real \(x\). So the sign of the rational expression \(\frac{x^3 - 4x}{x^2 + 2}\) is the same as the sign of the numerator \(x^3 - 4x\).

Step2: Factor the numerator

Factor the numerator \(x^3 - 4x\):
\[

$$\begin{align*} x^3 - 4x&=x(x^2 - 4)\\ &=x(x - 2)(x + 2) \end{align*}$$

\]

Step3: Find critical points

Set the numerator equal to zero: \(x(x - 2)(x + 2)=0\). The critical points are \(x=-2\), \(x = 0\), and \(x = 2\). These points divide the real number line into four intervals: \((-\infty, -2)\), \((-2, 0)\), \((0, 2)\), and \((2, \infty)\).

Step4: Test intervals

• For \(x\in(-\infty, -2)\), let's take \(x=-3\). Then \(x(x - 2)(x + 2)=(-3)(-5)(-1)=-15<0\).
• For \(x\in(-2, 0)\), let's take \(x=-1\). Then \(x(x - 2)(x + 2)=(-1)(-3)(1)=3>0\).
• For \(x\in(0, 2)\), let's take \(x = 1\). Then \(x(x - 2)(x + 2)=(1)(-1)(3)=-3<0\).
• For \(x\in(2, \infty)\), let's take \(x = 3\). Then \(x(x - 2)(x + 2)=(3)(1)(5)=15>0\).

We also need to consider the critical points themselves. When \(x=-2\), \(x = 0\), or \(x = 2\), the numerator is zero, so the expression is zero, which satisfies \(\frac{x^3 - 4x}{x^2 + 2}\leq0\).

Step5: Determine the solution set

From the tests, the expression is less than or equal to zero when \(x\leq - 2\) (where it's negative or zero) or \(0\leq x\leq2\) (where it's negative or zero).

Answer:

x ≤ -2 or 0 ≤ x ≤ 2 (the option: x ≤ -2 or 0 ≤ x ≤ 2)