如图 。。。求极值
1个回答
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f(x)=(2/3)x-x^(2/3)
f'(x)=2/3-(2/3)x^(-1/3)
f'(x)=0
2/3-(2/3)x^(-1/3)=0
x^(-1/3)=1
x=1
f'(x)>0
2/3-(2/3)x^(-1/3)>0
x^(-1/3)<1
x^(1/3)>1
x>1
f'(x)<0
x<1
f(x)在x=1处取得极小值:
f(1)=(2/3)×1-1^(2/3)
=2/3-1
=-1/3
f'(x)=2/3-(2/3)x^(-1/3)
f'(x)=0
2/3-(2/3)x^(-1/3)=0
x^(-1/3)=1
x=1
f'(x)>0
2/3-(2/3)x^(-1/3)>0
x^(-1/3)<1
x^(1/3)>1
x>1
f'(x)<0
x<1
f(x)在x=1处取得极小值:
f(1)=(2/3)×1-1^(2/3)
=2/3-1
=-1/3
追问
还有x=0时的情况呀,,我就是不太清楚这个(๑`・ᴗ・´๑)
追答
f(x)在x<1时,单调递减,在x=0处不可能取得极值。
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