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Critical Point

Solve the fol...

Question

Solve the following inequality:
$\frac{2 x}{x^{2} - 9} ⩽ \frac{1}{x + 2}$

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Solution

$\frac{2 x}{x^{2} - 9} \leq \frac{1}{x + 2} \frac{2 x}{x^{2} - 9} - \frac{1}{x + 2} \leq 0 \frac{2 x^{2} + 4 x - x^{2} + 9}{(x + 2) (x^{2} - 9)} \leq 0 \frac{2 x^{2} + 4 x - x^{2} + 9}{(x + 2) (x - 3) (x + 3)} \leq 0$
As discriminant of $x^{2} + 4 x + 9 < 0$ for every $x \in R, x^{2} + 4 x + 9 > 0 \forall x \in R$
$\frac{1}{(x + 2) (x - 3) (x + 3)} \leq 0 x \in (- \infty, - 3) \cup (- 2, 3)$

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