用函數(shù)的導(dǎo)數(shù)判定函數(shù)單調(diào)性的法則
(1)如果在(a,b)內(nèi),f′(x)>0,則f(x)在此區(qū)間是增函數(shù),(a,b)為f(x)的單調(diào)增區(qū)間;
(2)如果在(a,b)內(nèi),f′(x)<0,則f(x)在此區(qū)間是減函數(shù),(a,b)為f(x)的單調(diào)減區(qū)間.
1.判斷(正確的打“√”,錯(cuò)誤的打“×”)
(1)函數(shù)f(x)在定義域上都有f′(x)>0,則函數(shù)f(x)在定義域上單調(diào)遞增. ( )
(2)函數(shù)在某一點(diǎn)的導(dǎo)數(shù)越大,函數(shù)在該點(diǎn)處的切線越“陡峭”. ( )
(3)函數(shù)在某個(gè)區(qū)間上變化越快,函數(shù)在這個(gè)區(qū)間上導(dǎo)數(shù)的絕對(duì)值越大. ( )
[答案] (1)× (2)× (3)√
2.函數(shù)y=f(x)的圖象如圖所示,則( )
A.f′(3)>0
B.f′(3)<0
C.f′(3)=0
D.f′(3)的正負(fù)不確定
[解析] 由圖象可知,函數(shù)f(x)在(1,5)上單調(diào)遞減,則在(1,5)上有f′(x)<0,故f′(3)<0.
[答案] B
3.已知函數(shù)f(x)=x2-x,則f(x)的單調(diào)遞增區(qū)間為________.
[解析] ∵f′(x)=x-1,令f′(x)>0,解得x>1,
故f(x)的單調(diào)遞增區(qū)間是(1,+∞).
[答案] (1,+∞)