Question

不等式题

已知函数f(x)=ax³-cx,x∈[-1,1]
（1）若a=4,c=3,求证：对任意x∈[-1,1]，恒有|f(x)|≤1;
（2）若对任意x∈[-1,1]，恒有|f(x)|≤1，求证：|a|≤4
怎么做，为什么这么做？帮忙解析一下

Answer 1

应该是学了导数再做吧
(1)y=4x^3-3x, 求导数y'=12x^2-3=3(4x^2-1)，令y'=0,可得极值点x=-1/2或x=1/2
画表分析可得函数在x=-1/2取极大值,x=1/2取极小值,
同时可得 在定义域[-1,1]上，x=-1/2同时也是最大值点，x=1/2也是最小值点
即此时函数最大f(-1/2)=1, 最小f(1/2)=-1
所以|f(x)|≤1.
(2)也要用导数来做···求极值点 ，考虑极值点与[-1.1]端点的位置关系得最值.你自己试试吧.

Answer 2

1.
f(x)=4x^3-3x
f"(x)=12x^2-3
The absolute value of f(-1) f(-0.5) f(0.5) and f(1) are not lager than 1, which are possible extremum points.i.e the absolute value of f(x) is not lager than 1 when x∈[-1,1]∩R

2.
f'(x)=3ax^2-c
Let x be 1, then we get the inequality:
-1<a-c<1
Plainly it is equivalent to a-1<c<a+1.
Suppose a>4,then √(c/3a) is between 0 and 1.
Substitute x=±√(c/3a) to |f(x)|≤1 <=>
|x(ax^2-c)|≤1
=> |x(c/3-c)|=2/3|c*√(c/3a)|≤1
But then (2/3)(a-1)*√((a-1)/3a)<2/3|c*√(c/3a)|≤1
<=>
(a-1)^3/(3a)<4/9
<=>
(a-1)^3<(4/3)a
<=>
(a-1)^2*(1-1/a)<4/3
<=>
(4-1)^2*(1-1/4)<(a-1)^2*(1-1/a)<4/3
A contradiction.
The proof is the same for a<-4.
(Really His Mother Egg Hurt!)