求函數(shù)f(x)=(x-1)2(x+1)3的極值與單調(diào)區(qū)間.

數(shù)學(xué)人氣：148 ℃時間：2020-04-15 03:24:56

優(yōu)質(zhì)解答

f(x)=(x-1)^2 * (x+1)^3
f'(x) = (x-1)^2 * 3(x+1)^2 + (x+1)^3 * 2(x-1) = (x+1)^2 * (x-1) * { 3(x-1)+2(x+1) }
= (x+1)^2 * (x-1) * (5x-1)
= 5(x+1)^2 * (x-1/5) * (x-1)
x＜1/5或x＞1時,f'(x)＞0,f(x)單調(diào)增；1/5＜x＜1時,f'(x)＜0,f(x)單調(diào)減少
∴單調(diào)增區(qū)間：（-無窮大,1/5）U（1,+無窮大）；單調(diào)減區(qū)間（1/5,1)
x=1/5時有極大值f(1/5) = (1/5-1)^2 * (1/5+1)^3 = 3456/3125
x=1時有極小值f(1) = 0