导数综合题-基础篇

分离参量并猜根
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已知函数单调性求参数范围

1.若函数f(x)是定义域D内某个区间I上的增函数,且F(x)=f(x)x在I上是减函数,则    称函数f(x)是I上的“单反减函数”.已知f(x)=ln⁡x,g(x)=2x+2x+aln⁡x(a∈R).      (1)判断函数f(x)在(0,1)内是不是“单反减函数”;                                                               (2)若函数g(x)是[1,+∞)上的“单反减函数”,求实数a的取值范围.                                   1.若函数 f(x) 是定义域 D 内某个区间 I 上的增函数, 且 F(x)=\frac{f(x)}{x} 在 I 上是减函数, 则~~~~\\称函数 f(x) 是 I 上的 “单反减函数”. 已知 f(x)=\ln x , g(x)=2 x+\frac{2}{x}+a \ln x(a \in \mathbf{R}) .~~~~~~\\ (1) 判断函数 f(x) 在 (0,1) 内是不是 “单反减函数”;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2) 若函数 g(x) 是 [1,+\infty) 上的 “单反减函数”,求实数 a 的取值范围.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案

(1)不是单反减函数（2）0≤a≤4                                                                                        (1)不是单反减函数（2）0\le a\le4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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放缩一个量证明不等式

2.函数f(x)=ex−2x+2a.                                                                                                     求证：当a>ln2−1且x>0时，ex>x2−2ax+1.                                                          2.函数f(x)=e^x-2x+2a.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\求证：当a>ln2-1且x>0时，e^x>x^2-2ax+1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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练习1：已知f(x)=ax2+x−1ex,求证：当a≥1时，f(x)+e≥0.                                  练习1：已知f(x)=\frac{ax^2+x-1}{e^x} ,求证：当a\ge 1时，f(x)+e\ge 0.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案

只需证明a=1时，f(x)+e≥0                                                                                             只需证明a=1时，f(x)+e\ge 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
练习2：当x>0时，证明：ex−sin⁡x−1>xln⁡x.                                                             练习2：当x>0时，证明：e^x-\sin x-1>x\ln x.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案

提示：先放缩,当x>0时,x>sin⁡x.只需证ex−x−1>xln⁡x,令f(x)=ex−x−1−  xln⁡x,f′(x)=ex−ln⁡x−2,f′′(x)=ex−1x,f′′(12)=e−2<0,f′′(1)>0,∴∃x0∈(12,1),即ex0=1x0,f′′(x0)=0,∴f′(x)的极小值为f′(x0)=ex0−ln⁡x0−2=                  x0+1x0−2>0,∴f′(x)≥f′(x0)>0,∴f(x)在(0,+∞)单增，f(x)>f(0)=              lim⁡x→0xln⁡x=lim⁡x→0(ln⁡x)′(1x)′=lim⁡x→0x=0,∴f(x)>0.                                                                  提示：先放缩,当x>0时,x>\sin x.只需证e^x-x-1>x\ln x,令f(x)=e^x-x-1-~~\\x\ln x,f'(x)=e^x-\ln x-2,f''(x)=e^x-\frac{1}{x},f''(\frac{1}{2})=\sqrt{e}-2<0,f''(1)>0,\therefore \exists x_0\in\\ (\frac{1}{2},1) ,即e^{x_0}=\frac{1}{x_0} ,f''(x_0)=0,\therefore f'(x)的极小值为f'(x_0)=e^{x_0}-\ln{x_0}-2=~~~~~~~~~~~~~~~~~~\\x_0+\frac{1}{x_0}-2>0,\therefore f'(x)\ge f'(x_0)>0,\therefore f(x)在(0,+\infty )单增，f(x)>f(0)=~~~~~~~~~~~~~~\\\lim_{x \to 0} x\ln x=\lim_{x \to 0} \frac{(\ln x)'}{(\frac{1}{x})' } =\lim_{x \to 0} x=0,\therefore f(x)>0.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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练习3：已知函数f(x)=aex−ln⁡x−1,证明：当a≥1e时，f(x)≥0.                             练习3：已知函数f(x)=ae^x-\ln x-1,证明：当a\ge\frac{1}{e}时，f(x)\ge0.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案

提示：只需证明a=1e时，f(x)≥0                                                                                       提示：只需证明a=\frac{1}{e}时，f(x)\ge 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

分离参量并猜根

3.设f(x)=ax+axln⁡x,a>0.若对于∀x∈[12,+∞),有f(x)≤e2x恒成立，求a的         最大值.                                                                                                                                       3.设f(x)=ax+ax\ln x,a>0.若对于\forall x\in [\frac{1}{2},+\infty ),有f(x)\le e^{2x}恒成立，求a的~~~~~~~~~\\最大值.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案

a≤e2,所以a的最大值为e2.                                                                                                     a\le e^2,所以a的最大值为e^2.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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