【高中函数问题】已知函数F(x)=-1/4x^4+ax^3+(a^2+5a-2)x^2/2+b （a、b为常数） ..1.当a=1时,F(x)=0有两个不相等的实根谋求b的范围2.若F(x)有三个不同的即致电0,x1,x2.a为何值时,能使函数F(x)在x1（或x2）处

【高中函数问题】已知函数F(x)=-1/4x^4+ax^3+(a^2+5a-2)x^2/2+b （a、b为常数） ..1.当a=1时,F(x)=0有两个不相等的实根谋求b的范围2.若F(x)有三个不同的即致电0,x1,x2.a为何值时,能使函数F(x)在x1（或x2）处
【高中函数问题】已知函数F(x)=-1/4x^4+ax^3+(a^2+5a-2)x^2/2+b （a、b为常数） ..
1.当a=1时,F(x)=0有两个不相等的实根谋求b的范围
2.若F(x)有三个不同的即致电0,x1,x2.a为何值时,能使函数F(x)在x1（或x2）处取得的极值为b
3.若对任意a∈[-1,0]不等式F(x)>=-8在[-2,2]上恒成立,求b的范围

【高中函数问题】已知函数F(x)=-1/4x^4+ax^3+(a^2+5a-2)x^2/2+b （a、b为常数） ..1.当a=1时,F(x)=0有两个不相等的实根谋求b的范围2.若F(x)有三个不同的即致电0,x1,x2.a为何值时,能使函数F(x)在x1（或x2）处
1.
当a=1时
F(x)=-1/4x^4+x^3+(1^2+5-2)x^2/2+b
＝-1/4x^4+x^3+2x^2+b
求导得：
F '(x)＝-x^3+3x^2+4x
＝-x（x^2-3x-4）
＝-x（x-4）（x+1）
令F '(x)＞0得：
x∈（-∞,-1）∪（0,4）
令F '(x)＜0得：
x∈（-1,0）∪（4,+∞）
∴F(x)于（-∞,-1）,（0,4）↗
于（-1,0）,（4,+∞）↘
∵F(x)=0有两个不相等的实根
∴有：
情况①：
F(0)＞0
即b＞0
情况②：
F(-1)＞0＞F(4)
即¼-1+2+b＞0＞4³+4³+2×4²+b不成立
情况③：
F(-1)＜0＜F(4)
即¼-1+2+b＜0＜4³+4³+2×4²+b恒成立
综上,b＞0
2.
求导该函数,得导函数为
F '(x)＝-x[x^2-3ax-(a^2+5a-2)]
＝-x[x^2-3ax-(a^2+5a-2)]
∵F(x)有三个不同的即致电0,x1,x2
∴F '(x1)＝F '(x2)＝0
即x1^2-3ax1-(a^2+5a-2)＝x2^2-3ax2-(a^2+5a-2)＝0⑴
∴若F(x1)=x1²[-1/4x1^2+ax1+(a^2+5a-2)/2]+b＝b
即x1²[-1/4x1^2+ax1+(a^2+5a-2)/2]＝0
即[-1/4x1^2+ax1+(a^2+5a-2)/2]＝0
即x1^2-4ax1-2(a^2+5a-2)＝0⑵
⑴-⑵得
ax1+a^2+5a-2＝0
x1＝a+5-2/a（a≠0）
将x1＝a+5-2/a代入⑴得并且化简后解得（化简过程有点繁琐,若楼主仍然需要过程可在追问中追问我）：
（a-1）［a-（2/3）］（-3a²-15a+6）＝0
a1＝1,a2＝2/3,a3＝(-5-√33)/2,a4＝(-5+√33)/2
3.
求导该函数,得导函数为
F '(x)＝-x[x^2-3ax-(a^2+5a-2)]
由于括号“［］”中的函数的判别式
△=9a^2+4(a^2+5a-2)
＝11a^2+20a-8
在a∈[-1,0]时△<0恒成立
∴x^2-3ax-(a^2+5a-2)＞0恒成立
∴原函数的导数在x∈[-2,0]时为+,x∈[0,2]时为-
即x=0时原函数取最小值.\x0d
∴F(x)min=F(0)=b≥-8\x0d
所以b≥-8
楼主啊,这可是绝世烦题啊!＋点悬赏分吧≥﹏≤

题目上是不是不太清楚？

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