求拐点
曲线y=(x-1)(x-2)^2(x-3)^3(x-4)^4的拐点是??
求导 lny=ln(x-1)+2ln(x-2)+3ln(x-3)+4ln(x-4) y'/y=1/(x-1)+2/(x-2)+3/(x-3)+4/(x-4) y'=(x-1)(x-2)^2(x-3)^3(x-4)^4[1/(x-1)+2/(x-2)+3/(x-3)+4/(x-4)] =(x-1)(x-2)^2(x-3)^3(x-4)^4[A/(x-1)(x-2)(x-3)(x-4)] A就是通分后的分子,不去管它 y'=A/(x-2)(x-3)^2(x-4)^3 再求导的话y''=AB/(x-3)(x-4)^2 这样y''=0的话就有x=3或x=4 但是x=4时左右极限不一样, 所以拐点 x=3
答:y=[(x-1)(x-3)]^2 y'=2(x-1)(x-3)(x-3+x-2)=4(x-1)(x-2)(x-3) =4(x^3-6x^2+11x-6) y"=...详情>>
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