Question

已知函数f（x）=(4x2-7)/(2-x) x属于[0,1] 求单调区间

要过程详细

Answer 1

函数在[0,1/2]上单调减；在[1/2,1]上单调增

Answer 2

f(x)=ln(2-x) ax, 定义域:2-x>0, x<2
∴f'(x)=-1/(2-x) a=a-1/(2-x)
令f'(x)>=0, 则a>=1/(2-x), 2-x>=1/a, ∴x<=2-1/a
令f'(x)<=0, 则a<=1/(2-x), 2-x<=1/a, ∴x>=2-1/a
综上,单调增区间为(-∞,2-1/a],单调减区间为[2-1/a,2)

Answer 3

你题目都不明白...什么是4x2-7？