2个回答
展开全部
2x^3-3x^2-6x+2=0,
f(-2) = -14, f(-1) = 3, f(-1/2)= 4, f(0) = 2, f(1/2) = -3/2, f(1) = -5, f(2) = -6, f(3) = 11,
f(x) = 2x^3 - 3x^2 -6x+2,
f'(x) = 6x^2-6x-6 = 6[(x-1/2)^2 - 5/4],
故(0, 1/2)上存在一点a, 满足f(a) = 0.,
(2,3)上存在一点b, 满足f(b) = 0,
(-2,-1)上存在一点c, 满足f(c) = 0,
f(-2) = -14, f(-1) = 3, f(-1/2)= 4, f(0) = 2, f(1/2) = -3/2, f(1) = -5, f(2) = -6, f(3) = 11,
f(x) = 2x^3 - 3x^2 -6x+2,
f'(x) = 6x^2-6x-6 = 6[(x-1/2)^2 - 5/4],
故(0, 1/2)上存在一点a, 满足f(a) = 0.,
(2,3)上存在一点b, 满足f(b) = 0,
(-2,-1)上存在一点c, 满足f(c) = 0,
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
