Question

设可导函数f(x)满足：f'(x)+xf^2(x)=x^2,且f(0)=0则？

A x=0为f(x)的极大值点B x=0为f(x)的极小值点C (0,0)为f(x)的拐点D x=0不是f(x)的极值点，(0,0)也不是f(x)的拐点

Answer 1

ans: C

f(0) =0

f'(x) +x[f(x)]^2 =x^2

x=0, =>f'(0) =0         绝和          碰宏返  (1)

f'(x) +x[f(x)]^2 =x^2

两边求导

f''(x) + [f(x)]^2 + 2xf(x).f'(x) = 2x

x=0

f''(0) + [f(0)]^2 + 0 = 0

=> f''(0) =0                                (2)

f''(x) + [f(x)]^2 + 2xf(x).f'(x) = 2x

两边求导

f''笑饥'(x) + 2f(x).f'(x)  + 2{f(x).f'(x) + x[f'(x)]^2  + xf(x)f''(x) } = 2

x=0

f'''(0) + 2f(0).f'(0)  + 2{f(0).f'(0) + 0  + 0 } = 2

f'''(0) =2 ≠0                                    (3)

from (1) and (2) and (3)

x=0 是拐点