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# 微積分期中考的題目..因為有絕對值~我寫錯了!請教我一下!!

Given f(x)=│x^3-3x│

1.Determine the 臨界點(critical points)of f

2.Determine all the extrema

3.Is there any points of inflection(反曲點)

4.Identify where f is increasing,decreasing, concave up ,concave down(上凹&下凹)?

分成四種情況來討論...

那假如要畫出它的圖的話要怎麼畫呢?

### 1 Answer

- ?Lv 71 decade agoFavorite Answer
f(x)=|x3-3x|當x3-3x<0,x(x2-3)<0,x>0且x2-3<0---(1)或x<0,x2-3>0----(2)由(1),0<x<√3由(2),x<-√3當x3-3x≧0,x(x2-3)≧0=>x≧0,x2-３≧0--(3)或x≦0,x2-3≦0---(4)由(3),x≧√3由(4)-√3≦x≦0故要分四種情況來討論:當x3-3x<0,x<-√3時1.f(x)=-x3+3x,f'(x)=-3x2+3,f"(x)=-6x,當f'(x)=0,x=±1不在x的區間之內,∴f(x)沒有臨界點,又f'(x)<0 on (-∞,-√3),f is deacreasing on (-∞,-√3),f has no extream,no inflection point2.當x3-3x<0,0<x<√3f'(x)=-3x2+3,f"(x)=-6x,令f'(x)=0=>x=±1,x=1 is critical point,and f"(1)=-6<0,f has local maximu at x=1,it's value is f(1)=2For 0<x≦1,f'(x)≧0,is increasing for1<x<√3,f'(x)<0,is deacreasing so f is concave down on (0,√3),f has no inflection point3.當x3-3x≧0,x≧√3,f'(x)=3x2-3,f"(x)=6xLet f'(x)=0=>x=±1 is not on interval [√3,∞),so f has no critical pointf'(x)>0,on [√3,∞),so f is increasing on [√3,∞),f has minumum at x=√3,it's value is f(√3)=0f has no inflection point4.When x3-3x≧0,-√3≦x≦0,f'(x)=3x2-3.f"(x)=6x,Let f'(x)=3x2-3=0=>x=-1 so -1 is a critical point of f=>f"(-1)<0,f has local maximum at x=-1f'(x)≧0 on [-√3,1] so is increasing on [-√3,1] f'(x)≦0 on[-1,0]so is deacreasing on [-1,0]Hence f is concave down on [-√3,1],f has no inflection point

2006-05-01 17:15:46 補充：

這就變成把四種情況的圖形通通畫出來,不過圖形是大同小異,只是差在遞增,遞減的部分會不一樣

Source(s): me