Question

Найти критические точки функции f(x)= 3x^2-6x+1

Answer 1

$y=3 x^{2} -6x+1\\ y=0\iff 3 x^{2} -6x+1=0\\ \Delta=(-6)^2-4*3*1=36-12=24=4*6\\ x_1= \frac{6-2 \sqrt{6} }{6} = \frac{3- \sqrt{6} }{3} \\ x_2= \frac{3+ \sqrt{6} }{3} \\ x_0= \frac{6}{2*3} =1\\ y_{MIN}=3*1^2-6*1+1=3-6+1=4-6=-2\\ MIN(1;-2)\\$