也就是求其導(dǎo)數(shù)的零點(diǎn).
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,可得到三個(gè)x的值,所以有三個(gè)極值點(diǎn).這樣算好復(fù)雜的說(shuō)，選擇題有沒(méi)有更簡(jiǎn)單一點(diǎn)的辦法呢、？謝謝咯要是選擇題，比最高指數(shù)大那些選項(xiàng)都要去掉，至于具體的還得計(jì)算，計(jì)算過(guò)程中可能含有1個(gè)公因式，則數(shù)量又要減去1個(gè)。本題就是這樣，后面含有公因式(x-2).