Question

Решите неравенство: f′(х)≥0,  f(х)=х3+3х4 -3x2+1.​

Answer 1

Ответ:

$f(x)=x^3+3x^4-3x^2+1\\\\f'(x)=3x^2+12x^3-6x=3x\, (4x^2+x-2)=3x\Big (x-\dfrac{1-\sqrt{33}}{2}\Big)\Big(x-\dfrac{1+\sqrt{33}}{2}\Big)\\\\4x^2+x-2=0\ \ ,\ \ D=1+4\cdot 4\cdot 2=33\ ,\\\\x_1=\dfrac{1-\sqrt{33}}{2}\approx -2,37\ \ ,\ \ \ \ x_2=\dfrac{1+\sqrt{33}}{2}\approx 3,37\\\\\3x\Big (x-\dfrac{1-\sqrt{33}}{2}\Big)\Big(x-\dfrac{1+\sqrt{33}}{2}\Big)\geq 0\\\\\\znaki\ f'(x):\ \ ---\Big (\dfrac{1-\sqrt{33}}{2}\Big)+++(\ 0\ )---\Big(\dfrac{1+\sqrt{33}}{2}\Big)+++$

$x\in \Big [\, \dfrac{1-\sqrt{33}}{2}\ ;\ 0\ \Big]\cup \Big [\, \dfrac{1+\sqrt{33}}{2}\ ;+\infty \, \Big)$