Question

Решить неравенство:
f ' (x) > 0, если f(x) = 9х3 + 3х2

Answer 1

$f(x) = 9x^{3} + 3x^{2}$

$f'(x) = 9 \cdot 3x^{2} + 3 \cdot 2x = 27x^{2} + 6x$

Решим неравенство $f'(x) > 0$, то есть $27x^{2} + 6x > 0$

$27x^{2} + 6x > 0$

$3x(9x + 2) > 0$

$\left[\begin{array}{ccc}\displaystyle \left \{ {{3x > 0, \ \ \ \ \, } \atop {9x + 2 > 0}} \right. \\\displaystyle \left \{ {{3x < 0, \ \ \ \ \, } \atop {9x + 2 < 0}} \right.\\\end{array}\right \ \ \ \ \ \ \ \left[\begin{array}{ccc}\displaystyle \left \{ {{x > 0, \ \ \, } \atop {x > -\dfrac{2}{9} }} \right. \\\displaystyle \left \{ {{x < 0, \ \ \, } \atop {x < -\dfrac{2}{9}}} \right.\\\end{array}\right \ \ \ \ \ \left[\begin{array}{ccc}x > 0, \ \ \\x < -\dfrac{2}{9} \\\end{array}\right$

$x \in \left(-\infty; \ -\dfrac{2}{9} \right) \cup (0; \ +\infty)$

Ответ: $x \in \left(-\infty; \ -\dfrac{2}{9} \right) \cup (0; \ +\infty)$