Question

Sa se rezolve inceuatia: $x^{2} \leq \frac{16}{x}$

Answer 1

Intai tinem cont ca nu are sens $\cfrac{16}{0}$, deci x nu poate fi 0.


$x^2\le\cfrac{16}{x}\ \Bigr| \times x\implies x^3\le16\implies x\le\sqrt[\scriptstyle 3]{16}$

$\implies\boxed{ x\in\left(-\infty,\ \sqrt[\scriptstyle 3]{16}\right]\setminus \{0\}}$

Answer 2

$\it x^2\leq\dfrac{16}{x} \Rightarrow \dfrac{x^2}{16}\leq \dfrac{1}{x} \Rightarrow \dfrac{1}{x}\geq\dfrac{x^2}{16}\geq0 \Rightarrow x > 0\ \ \ \ \ (1)\\ \\ \\ x^2\leq\dfrac{16}{x}\Big|_{\cdot x} \Rightarrow x^3\leq16x^2\leq\dfrac{16}{x} \Rightarrow x\leq\sqrt[3]{\it 16}\ \ \ \ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow x\in(0,\ \ \sqrt[3]{\it16}]$