设函数f(x)=x(x-1)(x-a),(a>1)
(1)求导数f'(x);并证明f(x)有两个不同极值点x1、x2(2)若不等式f(x1)+f(x2)<=0成立,求a的取值范围...
(1)求导数f'(x);并证明f(x)有两个不同极值点x1、x2
(2)若不等式f(x1)+f(x2)<=0成立,求a的取值范围 展开
(2)若不等式f(x1)+f(x2)<=0成立,求a的取值范围 展开
2个回答
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f'(x)=(x-1)(x-a)+x(x-a)+x(x-1)
=x^2-x-ax+a+x^2-ax+x^2-x
=3x^2-2(a+1)x+a
令f'(x)=0,
3x^2-2(a+1)x+a=0,
x^2-2(a+1)x/3+a/3=0,
b^2-4ac=4(a+1)^2/9-4a/3=4(a^2-a+1)/9=4(a-1/2)^2/9+1/3>0,
所以f(x)有2个不同的极值点x1,x2.
x1+x2=(a+1)/3,x1*x2=a/3,
x1^+x2^2=(x1+x2)^2-2x1x2=(a+1)^2/9-2*a/3=(a^2-4a+1)/9,
x1^3+x2^3=(x1+x2)^3-3x1x2(x1+x2)=(a+1)^3/27-3*a/3*(a+1)/3=(a^3-6a^2-6a+1)/27,
f(x1)+f(x2)=x1(x1-1)(x1-a)+x2(x2-1)(x2-a)
=(x1^3+x2^3)-(a+1)(x1^2+x2^2)+a(x1+x2)
=(a^3-6a^2-6a+1)/27-(a+1)*(a^2-4a+1)/9+a*(a+1)/3
=(-2a^3+12a^2+12a-2)/27
要使f(x1)+f(x2)≤0,
-a^3+6a^2+6a-1≤0,
a^3-6a^2-6a+1≥0,
a^3+a^2-7a^2-7a+a+1≥0,
a^2(a+1)-7a(a+1)+(a+1)≥0,
(a+1)(a^2-7a+1)≥0,
(a+1)(a-7/2-3√5/2)(a-7/2+3√5/2)≥0,
-1≤a≤7/2-3√5/2或a≥7/2+3√5/2。
=x^2-x-ax+a+x^2-ax+x^2-x
=3x^2-2(a+1)x+a
令f'(x)=0,
3x^2-2(a+1)x+a=0,
x^2-2(a+1)x/3+a/3=0,
b^2-4ac=4(a+1)^2/9-4a/3=4(a^2-a+1)/9=4(a-1/2)^2/9+1/3>0,
所以f(x)有2个不同的极值点x1,x2.
x1+x2=(a+1)/3,x1*x2=a/3,
x1^+x2^2=(x1+x2)^2-2x1x2=(a+1)^2/9-2*a/3=(a^2-4a+1)/9,
x1^3+x2^3=(x1+x2)^3-3x1x2(x1+x2)=(a+1)^3/27-3*a/3*(a+1)/3=(a^3-6a^2-6a+1)/27,
f(x1)+f(x2)=x1(x1-1)(x1-a)+x2(x2-1)(x2-a)
=(x1^3+x2^3)-(a+1)(x1^2+x2^2)+a(x1+x2)
=(a^3-6a^2-6a+1)/27-(a+1)*(a^2-4a+1)/9+a*(a+1)/3
=(-2a^3+12a^2+12a-2)/27
要使f(x1)+f(x2)≤0,
-a^3+6a^2+6a-1≤0,
a^3-6a^2-6a+1≥0,
a^3+a^2-7a^2-7a+a+1≥0,
a^2(a+1)-7a(a+1)+(a+1)≥0,
(a+1)(a^2-7a+1)≥0,
(a+1)(a-7/2-3√5/2)(a-7/2+3√5/2)≥0,
-1≤a≤7/2-3√5/2或a≥7/2+3√5/2。
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