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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 − 3 ln a − 1 2 > 0 a^2-3\ln a-\dfrac 12>0 a2−3lna−21>0
\qquad 令 g ( a ) = a 2 − 3 ln a − 1 2 g(a)=a^2-3\ln a-\dfrac 12 g(a)=a2−3lna
−21,则 g ′ ( a ) = 2 a − 3 a g'(a)=2a-\dfrac 3a g′(a)=2a−a3
\qquad 当 a = 6 2 a=\dfrac{\sqrt 6}{2} a=26 时 g ′ ( a ) = 0 g'(a)=0 g′(a)=0
g ( a ) \qquad g(a) g(a)在 [ 6 2 , + ∞ ) [\dfrac{\sqrt 6}{2},+\infty) [26 ,+∞
)上单调递增,在 ( − ∞ , 6 2 ] (-\infty,\dfrac{\sqrt 6}{2}] (−∞,26 ]上单调递减
\qquad 所以 g ( a ) ≥ g ( 6 2 ) = 1 − ln 3 6 4 > 0 g(a)\geq g(\dfrac{\sqrt
6}{2})=1-\ln\dfrac{3\sqrt 6}{4}>0g(a)≥g(26 )=1−ln436 >0
\qquad 即 a 2 − 3 ln a − 1 2 > 0 a^2-3\ln a-\dfrac 12>0 a2−3lna−21>0
\qquad 由此可得 f ( x ) > 2 ln a + 3 2 f(x)>2\ln a+\dfrac 32 f(x)>2lna+23