为什么函数f(x)=(x-1)(x+1)²(x-2)³的极值点有三个?
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也就是求其导数的零点.
y'=(x+1)^2(x-2)^3+(x-1)*[2(x+1)(x-2)^3-(x+1)^2*3(x-2)^2]
=(x+1)^2(x-2)^3+2(x-1)(x+1)(x-2)^3-3(x-1)(x+1)^2(x-2)^2
=(x+1)^2(x-2)^3+(x+1)(x-2)^2[2(x-1)(x-2)-3(x-1)(x+1)]
=(x+1)^2(x-2)^3+(x+1)(x-2)^2(-x^2-6x+13)
=(x+1)(x-2)^2(x+1-x^2-6x+13)
=(x+1)(x-2)^2(-x^2-5x+14).
=-(x+1)(x-2)^2(x-2)(x+7).
=-(x+1)(x-2)^3(x+7).
令y'=0,可得到三个x的值,所以有三个极值点.
y'=(x+1)^2(x-2)^3+(x-1)*[2(x+1)(x-2)^3-(x+1)^2*3(x-2)^2]
=(x+1)^2(x-2)^3+2(x-1)(x+1)(x-2)^3-3(x-1)(x+1)^2(x-2)^2
=(x+1)^2(x-2)^3+(x+1)(x-2)^2[2(x-1)(x-2)-3(x-1)(x+1)]
=(x+1)^2(x-2)^3+(x+1)(x-2)^2(-x^2-6x+13)
=(x+1)(x-2)^2(x+1-x^2-6x+13)
=(x+1)(x-2)^2(-x^2-5x+14).
=-(x+1)(x-2)^2(x-2)(x+7).
=-(x+1)(x-2)^3(x+7).
令y'=0,可得到三个x的值,所以有三个极值点.
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