函数f(x)=x的3次方-3x+2在区间(1,2)内的函数值
A大于等于0
B小于等于0
C大于0
D小于0
人气:244 ℃ 时间:2019-10-19 00:38:08
解答
答:C
f(x)=x^3-3x+2
f'(x)=3x^2-3
解f'(x)=0得:x=-1或者x=1
x1时,f'(x)>0,f(x)单调递增
所以:f(x)在(1,2)区间上是单调递增
f(1)=1-3+2=0
f(2)=8-6+2=4
所以:f(x)>f(1)=0
选择Cf'(x)=3x^2-3 这一步怎么得出来的求导的结果没有学过求导==有不用求导的方法吗选择题,那就用代入特殊值排查
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