定义域x > 0
1.
f'(x) = mx - 2 + 1/x = (mx² - 2x + 1)/x
f'(1) = m - 1 = 0,m = 1
m = 1时,f'(x) = (x² - 2x + 1)/x = (x - 1)²/x
在x = 1两侧,f'(x)均> 0,x = 1不可能为函数fx的极值点
2.
(i) m = 0
f(x) = -2x + lnx
f'(x) = -2 + 1/x
f'(x) = 0,x = 1/2
0 < x < 1/2:f'(x) > 0,f(x)单调递增
(ii) m > 0
f'(x) = (mx² - 2x + 1)/x = [m(x - 1/m)² + 1 - 1/m]/x
(a) m ≥ 1
0 0,f'(x) ≥ 0
f(x)在定义域x> 0内单调递增
(b) 0 < m < 1
f'(x) = (mx² - 2x + 1)/x = 0
mx² - 2x + 1 = 0的解为x₁,₂ = [1 ±√(1-m)]/m
0 < x < [1 - √(1-m)]/m或x > [1 +√(1-m)]/m 时,f'(x) > 0,f(x)单调递增
3.
g(x)里没有t,题有问题.