已知函数f(x)=ax²－e^x(a∈R)(注:e是自然对数的底数)(Ι)当a=1时,试判断f(x)的单调性并给予证明（Ⅱ）若f（x）有两个极值点x1,x2（x1＜x2）（i）求实数a的取值范围；（ii）证明：-e／2＜f（x1）＜-1
问题描述：
已知函数f(x)=ax²－e^x(a∈R)(注:e是自然对数的底数)
(Ι)当a=1时,试判断f(x)的单调性并给予证明
（Ⅱ）若f（x）有两个极值点x1,x2（x1＜x2）（i）求实数a的取值范围；（ii）证明：-e／2＜f（x1）＜-1
答
(I)
f(x) = x²－e^x
f'(x) = 2x - e^x
f"(x) = 2 - e^x = 0
x = ln2
x = ln2时,f'(x)取极值,f'(ln2) = 2ln2 - 2 = 2(ln2 - 1) f'(x) f(x)为单调减函数
(II)
(i) f(x)有两个极值点,f'(x) = 2ax - e^x = 0有两个根
g(x) = e^x,g'(x) = e^x
h(x) = 2ax,h'(x) = 2a
假定二者相切,切点为P(p,e^p)
切线为:y - e^p = e^p(x - p)
切线过原点:-e^p = e^p(0 - p)
p = 1
切点为P(1,e)
此时切线斜率e = 2a,a = e/2
a e/2时,2ax - e^x = 0有两个根
(可以想象直线y = 2ax 从a = 0开始,逆时针绕原点旋转,a = e/2开始与g(x)相切,a > e/2时相交)
(ii)
显然0 f'(x) = 2ax - e^x
f(x)有两个极值点,0 f(1) f(0) = -1
f(1) = a - e > e/2 - e = -e/2
-e/2