Question

【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） 急急急急。。。

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

Answer 1

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。
∴F(x)min=F(0)=b≥-8
所以b≥-8

楼主啊，这可是绝世烦题啊！＋点悬赏分吧≥﹏≤

Answer 2

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Answer 3

题目上是不是不太清楚？

追问

能指出那点不清楚么。。。我是从作业上打的。。

追答

-1/4x^4是指四分之一的x的四次方还是四x的四次方分之一

追问

-1/4 x的四次方