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f '(x)= 3x^2-12x+9 當f '(x)=0時,3x^2-12x+9 =0 即(x-1)(x-3)=0 解得x1=1,x2 =3 當x 3,時,f '(x)>0 所以,當x=1時,原函數有極大值1;當x=3時原函數有極小值-3 -------------------- 梳理知識,幫助別人,愉悅自己. “數理無限”團隊歡迎你
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