Question

求函數(shù)f(x)=x³-3x²-9x+5的極值.

Answer 1

思路分析：由求函數(shù)的極值的方法先求其導(dǎo)數(shù),解方程f′(x)=0,分區(qū)間討論f′(x)的符號,進(jìn)而得函數(shù)f(x)的極值.

解：f′(x)=3x²-6x-9=3(x+1)(x-3),

令f′(x)=0,解得x₁=-1,x₂=3,

∴x＜-1時,f′(x)＞0,函數(shù)f(x)遞增；-1＜x＜3時,f′(x)＜0,函數(shù)f(x)遞減；x＞3時,f′(x)＞0,函數(shù)f(x)遞增.

∴f(x)_極大值=f(-1)=10；f(x)_極小值=f(3)=-22.