a-maths

2008-06-07 7:07 pm
試求函數f(x)=10x6-24x5+15x4的相對極值。

回答 (1)

2008-06-07 8:37 pm
✔ 最佳答案
f(x)=10x6-24x5+15x4
f'(x)=60x5-120x4+60x3
For f'(x)=0
60x3(x2-2x+1)=0
x3(x-1)2=0
x=0 or 1
Use first derivative test
x x<0 x=0 0<1 x=1 x>1
f'(x) - 0 + 0 +
∴x=0 is the min. and x=1 is pt. of reflexion(Amaths level does not require tgis)
Min point=(0,0)


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