已知函數(shù)f(x)=x3-ax2+bx+c.
(Ⅰ)若函數(shù)y=f(x)的圖象上存在點(diǎn)P,使P點(diǎn)處的切線與x軸平行,求實(shí)數(shù)a,b的關(guān)系式;
(Ⅱ)若函數(shù)f(x)在x=-1和x=3時取得極值,且其圖象與x軸有且只有3個交點(diǎn),求實(shí)數(shù)c的取值范圍.
分析:(Ⅰ)已知函數(shù)f(x)=x3-ax2+bx+c.函數(shù)y=f(x)的圖象上存在點(diǎn)P,使P點(diǎn)處的切線與x軸平行,f'(x)=3x2-2ax+b,即該方程有根.△=4a2-12b≥0,則易得a,b的關(guān)系式;
(Ⅱ)若函數(shù)f(x)在x=-1和x=3時取得極值,由已知可得x=-1和x=3是方程f'(x)=3x2-2ax+b=0的兩根,可以求得a,b,再根據(jù)圖象與x軸有且只有3個交點(diǎn),等價于極大值大于0且極小值小于0,則易求c的取值范圍.
解答:解:(Ⅰ)f'(x)=3x
2-2ax+b,設(shè)切點(diǎn)為P(x
0,y
0),
則曲線y=f(x)在點(diǎn)P處的切線的斜率k=f'(x
0)=3x
02-2ax
0+b,
由題意,知f'(x
0)=3x
02-2ax
0+b=0有解,
∴△=4a
2-12b≥0即a
2≥3b.
(Ⅱ)由已知可得x=-1和x=3是方程f'(x)=3x
2-2ax+b=0的兩根,
∴
-1+3=,
-1×3=,
∴a=3,b=-9.(7分)
∴f'(x)=3(x+1)(x-3),
∴f(x)在x=-1處取得極大值,在x=3處取得極小值.
∵函數(shù)y=f(x)的圖象與x軸有且只有3個交點(diǎn),∴
又f(x)=x
3-3x
2-9x+c,∴
,
解得-5<c<27.
點(diǎn)評:要明確導(dǎo)數(shù)的幾何意義,認(rèn)真讀題,了解其題意并結(jié)合函數(shù)圖象列出符合條件的不等式組,例如,若函數(shù)f(x)在x=-1和x=3時取得極值,且圖象與x軸有且只有3個交點(diǎn),等價于極大值大于0且極小值小于0,這點(diǎn)要結(jié)合函數(shù)圖象去理解.