Question

已知函數(shù)f(x)=x⁴+bx³+cx²+dx+e(x∈R)在x=0和x=1處取得極值.

(1)求d的值及b,c的關(guān)系式(用c表示b),并指出c的取值范圍;

(2)若函數(shù)f(x)在x=0處取得極大值.

①判斷c的取值范圍;

②若此時(shí)函數(shù)f(x)在x=1時(shí)取得最小值,求c的取值范圍.

Answer 1

解:(1)∵f′(x)=2x³+3bx²+2cx+d,

又∵f′(0)=f′(1)=0,

∴∴

∵f′(x)=2x³-2(c+1)x²+2cx,

即f′(x)=2x(x-1)(x-c),

∴c≠0且c≠1,即c的取值范圍是{c|c≠0且c≠1}.

(2)①∵f′(x)=2x(x-1)(x-c),

∴若c＜0.當(dāng)x∈(c,0)時(shí)f′(x)＞0,當(dāng)x∈(0,1)時(shí)f′(x)＜0,∴f(x)在x=0處取得極大值;

若0＜c＜1,當(dāng)x∈(-∞,0)時(shí)f′(x)＜0,當(dāng)x∈(0,c)時(shí)f′(x)＞0,∴f(x)在x=0處取得極小值;

若c＞1,當(dāng)x∈(-∞,0)時(shí)f′(x)＜0,當(dāng)x∈(0,1)時(shí)f′(x)＞0,∴f(x)在x=0處取得極小值.

綜上,若f(x)在x=0處取得極大值,則c的范圍為(-∞,0).10分

②當(dāng)c＜0時(shí),x∈(-∞,c)時(shí)f′(x)＜0,x∈(c,0)時(shí)f′(x)＞0,x∈(0,1)時(shí)f′(x)＜0,x∈(1,+∞)時(shí)f′(x)＞0,∴函數(shù)f(x)只能在x=c或x=1處取得最小值.要使f(x)在x=1處取得最小值,只要使得f(c)≥f(1).

∴c⁴+c³+e≥+c+e.

∴c⁴-2c³+2c-1≤0,即(c-1)³(c+1)≤0.

∴-1≤c＜0,即c的取值范圍是［-1,0).