Question

【导数题】已知:m∈R,函数f(x)=(x^2+mx+m)e^x.

(1)若函数f(x)没有零点，求m的范围
(2)若函数f(x)存在极大值，并记为g(m),求g(m)的表达式
(3)当m=0时，求证：f(x)≥x^2+x^3

Answer 1

e^x > 0
f(x) = (x^2+mx+m)e^x 无零点, 所以
x^2 + mx + m ≠ 0
或者说 x^2 + mx + m = 0 无解
判别式 m^2 - 4m < 0
m(m-4) < 0
0 < m < 4

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f(x)=(x^2+mx+m)e^x

f'(x) =
= (x^2 + mx + m)e^x + (2x + m) e^x
= [x^2 + (m+2)x + 2m]e^x

f''(x)
= [x^2 + (m+2)x + 2m]e^x + [2x + m+2]e^x
= [x^2 + (m+4)x + 3m+2]e^x

函数f(x)存在极大值 所以
f'(x) = 0 有解, 且在 有解处 f''(x) < 0
因为 e^x > 0 恒成立, 所以
x^2 + (m+2)x + 2m = 0 有解
(x + 2)(x+m) = 0
x1 = -2
x2 = -m

f''(-2) = [x^2 + (m+4)x + 3m+2]e^x
= [4 - 2(m+4) + 3m+2]/e^2
= (m-2)/e^2

f''(-m) = [m^2 - m(m+4) + 3m+2]/e^m
=(2 - m)/e^m

m < 2 时, f'(-2) = 0, f''(-2) < 0 , 在 x = -2 处存在极大值
g(m) = f(-2) = (x^2 + mx + m)e^x = (4 - 2m + m)/e^2 = (4 - m)/e^2

m > 2 时
f'(-m) = 0, f''(-m) < 0 , 在 x = -m 处存在极大值
g(m) = f(-m) = (x^2 + mx + m)e^x = (m^2 - m^2 + m)/e^m = m/e^m

综上所述
m < 2 时, g(m) = (4-m)/e^2
m > 2 时, g(m) = m/e^m

===================================
m = 0 时
f(x) = x^2 * e^x

求证：f(x)≥x^2+x^3 = x^2(x+1) 只需要证明
e^x > x + 1

设 h(x) = e^x - x - 1 , 现在要证明 h(x) 最小值 大于等于 0
h'(x) = e^x -1
另 h'(x) = 0 , 求 h(x) 极小值
e^x - 1 = 0
x = 0 处取极小值
h(0) = e^x - x - 1 = 1 - 0 - 1 = 0
而在 x < 0 一侧, h'(x) = e^x - 1 < 0
在 x > 0 一侧, h'(x) = e^x - 1 > 0
因此 x= 0 处的极小值 也是 h(x) 的最小值
h(x) = e^x - x - 1 ≥ 0
x^2 e^x ≥ x^2(x+1)
f(x)≥x^2 + x^3

Answer 2

e^x>0
f(x)=(x^2+mx+m)e^x无零点,所以
x^2+mx+m≠0
或者说x^2+mx+m=0无解
判别式m^2-4m<0
m(m-4)<0
0<m<4
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f(x)=(x^2+mx+m)e^x
f'(x)=
=(x^2+mx+m)e^x+(2x+m)e^x
=[x^2+(m+2)x+2m]e^x
f''(x)
=[x^2+(m+2)x+2m]e^x+[2x+m+2]e^x
=[x^2+(m+4)x+3m+2]e^x
函数f(x)存在极大值所以
f'(x)=0有解,且在有解处f''(x)<0
因为e^x>0恒成立,所以
x^2+(m+2)x+2m=0有解
(x+2)(x+m)=0
x1=-2
x2=-m
f''(-2)=[x^2+(m+4)x+3m+2]e^x
=[4-2(m+4)+3m+2]/e^2
=(m-2)/e^2
f''(-m)=[m^2-m(m+4)+3m+2]/e^m
=(2-m)/e^m
m<2时,f'(-2)=0,f''(-2)<0,在x=-2处存在极大值
g(m)=f(-2)=(x^2+mx+m)e^x=(4-2m+m)/e^2=(4-m)/e^2
m>2时
f'(-m)=0,f''(-m)<0,在x=-m处存在极大值
g(m)=f(-m)=(x^2+mx+m)e^x=(m^2-m^2+m)/e^m=m/e^m
综上所述
m<2时,g(m)=(4-m)/e^2
m>2时,g(m)=m/e^m
===================================
m=0时
f(x)=x^2*e^x
求证：f(x)≥x^2+x^3=x^2(x+1)只需要证明
e^x>x+1
设h(x)=e^x-x-1,现在要证明h(x)最小值大于等于0
h'(x)=e^x-1
另h'(x)=0,求h(x)极小值
e^x-1=0
x=0处取极小值
h(0)=e^x-x-1=1-0-1=0
而在x<0一侧,h'(x)=e^x-1<0
在x>0一侧,h'(x)=e^x-1>0
因此x=0处的极小值也是h(x)的最小值
h(x)=e^x-x-1≥0
x^2e^x≥x^2(x+1)
f(x)≥x^2+x^3