08 ce amaths

f(x)=x(x-6)²
f'(x)=x[2(x-6)]+(x-6)²
=(x-6)(2x+x-6)
=(x-6)(3x-6)
For f'(x)=0
x=6 or 2
f"(x)=3(x-6)+(3x-6)
=6x-24
f"(6)=12
f"(0)=-24
∴f(x) is max. when x=0 and is min. when x=6
∴Max. pt=(0,0), Min. pt. =(6,0)

2008-06-14 02:19:00 補充：
f'(x)=f'(x)
f"(x)=f"(x)
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f(x)=x(x^2-12x+36)
=x^3-12x^2+36x
df(x)/dx=3x^2-24x+36
=(3x-6)(x-6)
for (3x-6)(x-6)=0
x=2 or x=6
when x=2, f(x)=32
when x=6,f(x)=0
therefore, the maximum point is (2,32)
the minimum point is (6,0)