已知函数
f
(
x
)
=
a
(
e
x
+
a
)
−
x
f(x)=a(e^x+a)-x
f(x)=a(ex+a)−x
(1)讨论
f
(
x
)
f(x)
f(x)的单调性
(2)证明:当
a
>
0
a>0
a>0时,求证:
f
(
x
)
>
2
ln
a
+
3
2
f(x)>2\ln a+\dfrac 32
f(x)>2lna+23
解:
\quad
(1)
f
′
(
x
)
=
a
e
x
−
1
f'(x)=ae^x-1
f′(x)=aex−1
\qquad ① a > 0 a>0 a>0时, x = − ln a x=-\ln a x=−lna时 f ′ ( x ) = 0 f'(x)=0 f′(x)=0
f ( x ) \qquad f(x) f(x)在 [ − ln a , + ∞ ) [-\ln a,+\infty) [−lna,+∞)上单调递增,在 ( − ∞ , − ln a ] (-\infty,-\ln a] (−∞,−lna]上单调递减
\qquad ② a ≤ 0 a\leq 0 a≤0时, f ( x ) f(x) f(x)在 ( − ∞ , + ∞ ) (-\infty,+\infty) (−∞,+∞)上单调递减
\quad (2)由(1)得 x = − ln a x=-\ln a x=−lna时 f ( x ) f(x) f(x)取最小值
\qquad 题目即证 f ( − ln a ) > 2 ln a + 3 2 f(-\ln a)>2\ln a+\dfrac 32 f(−lna)>2lna+23
\qquad 即 1 + a 2 + ln a > 2 ln a + 3 2 1+a^2+\ln a>2\ln a+\dfrac 32 1+a2+lna>2lna+23, a 2 − ln a − 1 2 > 0 a^2-\ln a-\dfrac 12>0 a2−lna−21>0
\qquad 令 g ( a ) = a 2 − ln a − 1 2 g(a)=a^2-\ln a-\dfrac 12 g(a)=a2−lna−21,则 g ′ ( a ) = 2 a − 1 a g'(a)=2a-\dfrac 1a g′(a)=2a−a1
\qquad 当 a = 2 2 a=\dfrac{\sqrt 2}{2} a=22时 g ′ ( a ) = 0 g'(a)=0 g′(a)=0
g ( a ) \qquad g(a) g(a)在 [ 2 2 , + ∞ ) [\dfrac{\sqrt 2}{2},+\infty) [22,+∞)上单调递增,在 ( − ∞ , 2 2 ] (-\infty,\dfrac{\sqrt 2}{2}] (−∞,22]上单调递减
\qquad 所以 g ( a ) ≥ g ( 2 2 ) = − ln 2 2 = ln 2 > 0 g(a)\geq g(\dfrac{\sqrt 2}{2})=-\ln\dfrac{\sqrt 2}{2}=\ln\sqrt 2>0 g(a)≥g(22)=−ln22=ln2>0
\qquad 即 a 2 − ln a − 1 2 > 0 a^2-\ln a-\dfrac 12>0 a2−lna−21>0文章来源:https://www.toymoban.com/news/detail-496518.html
\qquad 得证 f ( x ) > 2 ln a + 3 2 f(x)>2\ln a+\dfrac 32 f(x)>2lna+23文章来源地址https://www.toymoban.com/news/detail-496518.html
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