函数f(x)=ax³+bx²+cx在区间[0,1]上是增函数,在区间(-∞,0),(1,+∞)上是减函数,f′(1/2)=3/2.(Ⅰ)求f(x)的解析式;(Ⅱ)与偶在区间[0,m](m>0)上恒有f(x)≤x成立,求m的取值范围.
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![函数f(x)=ax³+bx²+cx在区间[0,1]上是增函数,在区间(-∞,0),(1,+∞)上是减函数,f′(1/2)=3/2.(Ⅰ)求f(x)的解析式;(Ⅱ)与偶在区间[0,m](m>0)上恒有f(x)≤x成立,求m的取值范围.](/uploads/image/z/1331905-49-5.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3Dax%26%23179%3B%2Bbx%26%23178%3B%2Bcx%E5%9C%A8%E5%8C%BA%E9%97%B4%5B0%2C1%5D%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E5%9C%A8%E5%8C%BA%E9%97%B4%28-%E2%88%9E%2C0%29%2C%281%2C%2B%E2%88%9E%29%E4%B8%8A%E6%98%AF%E5%87%8F%E5%87%BD%E6%95%B0%2Cf%E2%80%B2%EF%BC%881%2F2%EF%BC%89%3D3%2F2.%EF%BC%88%E2%85%A0%EF%BC%89%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%EF%BC%9B%EF%BC%88%E2%85%A1%EF%BC%89%E4%B8%8E%E5%81%B6%E5%9C%A8%E5%8C%BA%E9%97%B4%5B0%2Cm%5D%EF%BC%88m%EF%BC%9E0%EF%BC%89%E4%B8%8A%E6%81%92%E6%9C%89f%EF%BC%88x%29%E2%89%A4x%E6%88%90%E7%AB%8B%2C%E6%B1%82m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.)
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函数f(x)=ax³+bx²+cx在区间[0,1]上是增函数,在区间(-∞,0),(1,+∞)上是减函数,f′(1/2)=3/2.(Ⅰ)求f(x)的解析式;(Ⅱ)与偶在区间[0,m](m>0)上恒有f(x)≤x成立,求m的取值范围.
函数f(x)=ax³+bx²+cx在区间[0,1]上是增函数,在区间(-∞,0),(1,+∞)上是减函数,f′(1/2)=3/2.
(Ⅰ)求f(x)的解析式;
(Ⅱ)与偶在区间[0,m](m>0)上恒有f(x)≤x成立,求m的取值范围.
函数f(x)=ax³+bx²+cx在区间[0,1]上是增函数,在区间(-∞,0),(1,+∞)上是减函数,f′(1/2)=3/2.(Ⅰ)求f(x)的解析式;(Ⅱ)与偶在区间[0,m](m>0)上恒有f(x)≤x成立,求m的取值范围.
答:
f(x)=ax³+bx²+cx在[0,1]上是增函数,
在(-∞,0),(1,+∞)上是减函数
则x=0和x=1是导函数f'(x)的零点:
f'(x)=3ax²+2bx+c
根据韦达定理有:
x1+x2=-2b/(3a)=1
x1*x2=c/(3a)=0
所以:c=0,b=-3a/2
f'(x)=3ax²-3ax
f'(1/2)=3a/4-3a/2=3/2
解得:a=-2,b=3,c=0
所以:f(x)=-2x³+3x²
2)
0=1或者x