若函数y=1/3x³-1/2ax²+(a-1)x+1在区间(1,4)内为减函数,在区间(6,+∞)内为增函数,求a的取值范围
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![若函数y=1/3x³-1/2ax²+(a-1)x+1在区间(1,4)内为减函数,在区间(6,+∞)内为增函数,求a的取值范围](/uploads/image/z/615410-26-0.jpg?t=%E8%8B%A5%E5%87%BD%E6%95%B0y%3D1%2F3x%26%23179%3B-1%2F2ax%26%23178%3B%2B%28a-1%29x%2B1%E5%9C%A8%E5%8C%BA%E9%97%B4%281%2C4%29%E5%86%85%E4%B8%BA%E5%87%8F%E5%87%BD%E6%95%B0%2C%E5%9C%A8%E5%8C%BA%E9%97%B4%EF%BC%886%2C%2B%E2%88%9E%EF%BC%89%E5%86%85%E4%B8%BA%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
若函数y=1/3x³-1/2ax²+(a-1)x+1在区间(1,4)内为减函数,在区间(6,+∞)内为增函数,求a的取值范围
若函数y=1/3x³-1/2ax²+(a-1)x+1在区间(1,4)内为减函数,
在区间(6,+∞)内为增函数,求a的取值范围
若函数y=1/3x³-1/2ax²+(a-1)x+1在区间(1,4)内为减函数,在区间(6,+∞)内为增函数,求a的取值范围
y=1/3x³-1/2ax²+(a-1)x+1
f'(x)=x²-ax+(a-1)=(x-1)[x-(a-1)]
(1)a-1≤1,
则x>1时,f'(x)>0
∴ f(x)在(1,+∞)上是增函数,与题意不符合.
(2)a-1>1, 即a>2时,
x