已知函數(shù)f(x)=x3+ax2+bx+c圖象上一點(diǎn)M(1,m)處的切線方程為y-2=0,其中a,b,c為常數(shù).
(1)當(dāng)a>-3時(shí),求函數(shù)f(x)的單調(diào)減區(qū)間(用a表示).
(2)若x=1不是函數(shù)f(x)的極值點(diǎn),求證:函數(shù)f(x)的圖象關(guān)于點(diǎn)M對(duì)稱(chēng).
【答案】
分析:(1)已知易得點(diǎn)M(1,m)在切線上,得到m=2,且切線斜率為0,列出相應(yīng)的等式,利用導(dǎo)數(shù)研究函數(shù)的單調(diào)區(qū)間的步驟求解.
(2)根據(jù)已知可以得出a,b,c的值,也就得到f(x),若證明f(x)的圖象關(guān)于點(diǎn)M對(duì)稱(chēng),只需證明f(x)的圖象上的任意一點(diǎn)P(x
,y
),關(guān)于點(diǎn)M的對(duì)稱(chēng)點(diǎn)Q(2-x
,4-y
)也在圖象上即可.
解答:解:(1)f(x)=x
3+ax
2+bx+c,f'(x)=3x
2+2ax+b,
由題意,知m=2,f(1)=1+a+b+c=2,f'(1)=3+2a+b=0,
即b=-2a-3,c=a+4.
當(dāng)a>-3時(shí),
,有
∴當(dāng)a>-3時(shí),函數(shù)f(x)的單調(diào)減區(qū)間為
(2)由(1)知:若x=1不是函數(shù)f(x)的極值點(diǎn),則
=1,
解出a=-3,b=3,c=1,f(x)=x
3-3x
2+3x+1=(x-1)
3+2.
設(shè)點(diǎn)P(x
,y
)是函數(shù)f(x)的圖象上任意一點(diǎn),則y
=f(x
)=(x
-1)
3+2,點(diǎn)P(x
,y
)關(guān)于點(diǎn)M(1,2)的對(duì)稱(chēng)點(diǎn)為Q(2-x
,4-y
),
∵f(2-x
)=(2-x
-1)
3+2=-(x
-1)
3+2=2-y
+2=4-y
,
∴點(diǎn)Q(2-x
,4-y
)在函數(shù)f(x)的圖象上.由點(diǎn)P的任意性知函數(shù)f(x)的圖象關(guān)于點(diǎn)M對(duì)稱(chēng).
點(diǎn)評(píng):本題考查導(dǎo)數(shù)的幾何意義,利用導(dǎo)數(shù)求解函數(shù)的單調(diào)區(qū)間及證明函數(shù)圖象關(guān)于點(diǎn)對(duì)稱(chēng)的方法,本題較好,是教學(xué)中的重點(diǎn)和難點(diǎn),同學(xué)們應(yīng)熟練掌握其方法步驟.