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函数(下)

2023-07-17 17:24 作者:24bs  | 我要投稿

f'(x) 定义域为 %5Cmathbb%20Rf(%5Cdfrac32-2x)%2Cf'(2%2Bx) 均为偶函数,判断以下说法正确性:

A. f(0)%3D0

B. g(-%5Cdfrac12)%3D0

C. f(-1)%3Df(4)

D. g(-1)%3Dg(2)

f(x)%5Crightarrow%20f(x%2B%5Cdfrac32)%5Crightarrow%20f(-2x%2B%5Cdfrac32),所以 f(x) 图像关于直线 x%3D%5Cdfrac32 对称;同理 f'(x)%0A 图像关于直线 x%3D2 对称。后略。总之选 BC。

a%3D0.1e%5E%7B0.1%7Db%3D%5Cdfrac19c%3D-%5Cln0.9,比较 a%2Cb%2Cc 大小关系。

c%3D-%5Cln%5Cdfrac9%7B10%7D%3D%5Cln%5Cdfrac%7B10%7D9%3D%5Cln(1%2B%5Cdfrac19)。由 %5Cln(x)%5Cle%20x-1 知 c%5Cle%20%5Cdfrac19%3Db。由于无法取等,故 c%3Cb

比较 a%2Cb%5CLeftrightarrow 比较 e%5E%7B0.1%7D%2C%5Cdfrac%7B10%7D9%5CLeftrightarrow 比较 0.1%2C-%5Cln(1-0.1)%5CLeftrightarrow 比较 -0.1%2C%5Cln(1-0.1)。由 %5Cln(1%2Bx)%5Cle%20x 得 %5Cln(1-0.1)%5Cle%20-0.1,即 %5Cdfrac%7B10%7D9%5Cge%20e%5E%7B0.1%7D,故 a%5Cle%20b,进而 a%3Cb

比较 a%2Cc%5CLeftrightarrow 比较 xe%5Ex%5Cmid_%7Bx%3D0.1%7D 和 -%5Cln(1-x)%5Cmid_%7Bx%3D0.1%7D。求导分析即可。总之 b%3Ea%3Ec

a%3D%5Cdfrac%7B31%7D%7B32%7D%2C%5C%20b%3D%5Ccos%5Cdfrac14%2C%5C%20c%3D4%5Csin%5Cdfrac14,比较 a%2Cb%2Cc 大小关系。

当 x%5Cin%20(0%2C%5Cdfrac%7B%5Cpi%7D2)%5Csin%20x%3Cx%3C%5Ctan%20x。作比得 c%3Eb。比较 a%2Cb%5CLeftrightarrow 比较 (1-%5Cdfrac%7Bx%5E2%7D%7B2%7D)%5Cmid_%7Bx%3D%5Cfrac14%7D 与 %5Ccos%20x%20%5Cmid_%7Bx%3D%5Cfrac14%7D。利用泰勒展式或求导即可。总之 a%3Cb%3Cc

9%5Em%3D10%2C%5C%20a%3D10%5Em-11%2C%5C%20b%3D8%5Em-9比较 a%2Cb%2C0 大小关系。

当 n%3E1 时,%5Clog_n%7B(n%2B1)%7D%3E%5Clog_%7Bn%2B1%7D%7B(n%2B2)%7D。后略。总之 a%3E0%3Eb

f(x)%3Dx%5E3%2Bax%5E2%2Bbx%2Bc0%5Cle%20f(-1)%3Df(-2)%3Df(-3)%5Cle%203,求 c 取值范围。

f(x)%3D(x%2B1)(x%2B2)(x%2B3)%2Bc-6。后略。


函数 f%3A%5Cmathbb%20R%5Crightarrow%20%5Cmathbb%20R,若负常数 a%2Cb 满足 %5Cforall%20x%5Cin%20%5Cmathbb%20Rf(x)%5Cle%20f(x%2Ba%2Bb)%2B2%5Csqrt%7Bab%7Df(x)%5Cle%20f(x-a-b)%2Ba%2Bb,证明:a%3Db

f(x)%5Cle%20f(x-a-b)%2Ba%2Bb%5Cle%20f(x-a-b)-2%5Csqrt%7Bab%7D,故 f(x%2Ba%2Bb)%5Cle%20f(x)-2%5Csqrt%7Bab%7D。又 f(x)%5Cle%20f(x%2Ba%2Bb)%2B2%5Csqrt%7Bab%7D,故由基本不等式取等条件即得 a%3Db


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