Question

已知函数f(x)=2x*3-6x*2+3,试判断函数零点的个数和存在区间

要完整过程

Answer 1

f(x)=2x^3-6x^2+ 3
f'(x) = 6x^2 - 12x =0
x(x-2) =0
x = 0 or 2
f''(x) = 12x-12
f''(0) = -12 ( max)
f''(2) = 12 ( min)

f(0) = 3 > 0
f(2) = 2(8)-6(4) + 3
= -5 < 0

f(x) is increasing on （-∞, 0]
from -ve -> +ve

f(x) is decreasing on [0, 2]
from +ve -> -ve

f(x) is increasing on [2, ∞)
from -ve -> +ve

f(-∞) = -∞

一个根属于（-∞, 0]
一个根属于[0, 2]
一个根属于[2, ∞)

total: 3 real roots

Answer 2

用几何画板一画就出来了