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帮忙解一道函数题,
已知一次函数f(x)=(m²-1)x+(m²-3m+2);
(1)若f(x)是减函数,且f(1)=0,求m的值
(2)若f(x)是奇函数,求m的值
人气:346 ℃ 时间:2020-07-31 15:26:36
解答
(1)因为f(1) = 0
所以 (m²-1)+(m²-3m+2) = 0
2m² - 3m + 1 = 0
(2m-1)(m-1) = 0
m= 1/2 或 m=1
当m = 1 时,m²-1 = 0 舍去
当m =1/2时 ,m²-1 = -3/4 满足f(x)是减函数
所以 m = 1/2
(2)因为f(x)是奇函数
所以m²-3m+2 = 0
(m-1)(m-2) = 0
m = 1 或 m = 2
当m = 1 时,m²-1 = 0 舍去
当m = 2 时 ,m²-1 = 3 满足条件
所以 m =2
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