考點(diǎn):二次函數(shù)的性質(zhì)
專題:函數(shù)的性質(zhì)及應(yīng)用
分析:求導(dǎo)函數(shù),對(duì)參數(shù)m進(jìn)行討論,即可確定函數(shù)f(x)的單調(diào)區(qū)間.
解答:
解:函數(shù)的定義域{x|x>0},f′(x)=mx+2+m
•=
令f′(x)=0,即mx
2+2x+m=0
(1)當(dāng)m=0時(shí),f′(x)>0,則函數(shù)在定義域上為增函數(shù);
(2)當(dāng)△=4(1-m
2)≤0,解得m≥1或m≤-1,
當(dāng)m≥1時(shí),f′(x)≥0,則函數(shù)在定義域上為增函數(shù),
當(dāng)m≤-1時(shí),f′(x)≤0,則函數(shù)在定義域上為減函數(shù);
(3)當(dāng)0<m<1時(shí),mx
2+2x+m=0兩根x
1,x
2(x
1<x
2)為負(fù)數(shù),所以函數(shù)在定義域上為增函數(shù)
當(dāng)-1<m<0時(shí),mx
2+2x+m=0兩根x
1,x
2(x
1<x
2)異號(hào)
x
1=
<0,x
2=
>0
函數(shù)f(x)在(0,
)上為增函數(shù),在(
,+∞)上為減函數(shù).
綜上所述:當(dāng)m≥0時(shí),函數(shù)y=f(x) 的增區(qū)間為(0,+∞);
當(dāng)m≤-1時(shí),函數(shù)y=f(x)的減區(qū)間(0,+∞);
當(dāng)-1<m<0時(shí),函數(shù)的增區(qū)間為(0,
),減區(qū)間為(
,+∞).
點(diǎn)評(píng):本題考查導(dǎo)數(shù)知識(shí)的運(yùn)用,考查函數(shù)的單調(diào)區(qū)間,屬于中檔題.