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函數(shù)(下)

2023-07-17 17:24 作者:24bs  | 我要投稿

f'(x)?定義域?yàn)?%5Cmathbb%20Rf(%5Cdfrac32-2x)%2Cf'(2%2Bx)?均為偶函數(shù),判斷以下說法正確性:

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)?圖像關(guān)于直線?x%3D%5Cdfrac32?對(duì)稱;同理?f'(x)%0A?圖像關(guān)于直線?x%3D2?對(duì)稱。后略??傊x BC。

a%3D0.1e%5E%7B0.1%7D,b%3D%5Cdfrac19c%3D-%5Cln0.9,比較?a%2Cb%2Cc?大小關(guān)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。由于無法取等,故?c%3Cb。

比較?a%2Cb%5CLeftrightarrow?比較?e%5E%7B0.1%7D%2C%5Cdfrac%7B10%7D9%5CLeftrightarrow?比較?0.1%2C-%5Cln(1-0.1)%5CLeftrightarrow?比較?-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,進(jìn)而?a%3Cb

比較?a%2Cc%5CLeftrightarrow?比較?xe%5Ex%5Cmid_%7Bx%3D0.1%7D?和?-%5Cln(1-x)%5Cmid_%7Bx%3D0.1%7D。求導(dǎo)分析即可??傊?b%3Ea%3Ec

a%3D%5Cdfrac%7B31%7D%7B32%7D%2C%5C%20b%3D%5Ccos%5Cdfrac14%2C%5C%20c%3D4%5Csin%5Cdfrac14,比較?a%2Cb%2Cc?大小關(guān)系。

當(dāng)?x%5Cin%20(0%2C%5Cdfrac%7B%5Cpi%7D2),%5Csin%20x%3Cx%3C%5Ctan%20x。作比得?c%3Eb。比較?a%2Cb%5CLeftrightarrow?比較?(1-%5Cdfrac%7Bx%5E2%7D%7B2%7D)%5Cmid_%7Bx%3D%5Cfrac14%7D?與?%5Ccos%20x%20%5Cmid_%7Bx%3D%5Cfrac14%7D。利用泰勒展式或求導(dǎo)即可??傊?a%3Cb%3Cc

9%5Em%3D10%2C%5C%20a%3D10%5Em-11%2C%5C%20b%3D8%5Em-9,比較?a%2Cb%2C0?大小關(guān)系。

當(dāng)?n%3E1?時(shí),%5Clog_n%7B(n%2B1)%7D%3E%5Clog_%7Bn%2B1%7D%7B(n%2B2)%7D。后略??傊?a%3E0%3Eb。

f(x)%3Dx%5E3%2Bax%5E2%2Bbx%2Bc,0%5Cle%20f(-1)%3Df(-2)%3Df(-3)%5Cle%203,求?c?取值范圍。

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


函數(shù)?f%3A%5Cmathbb%20R%5Crightarrow%20%5Cmathbb%20R,若負(fù)常數(shù)?a%2Cb?滿足?%5Cforall%20x%5Cin%20%5Cmathbb%20R,f(x)%5Cle%20f(x%2Ba%2Bb)%2B2%5Csqrt%7Bab%7Df(x)%5Cle%20f(x-a-b)%2Ba%2Bb,證明: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條件即得?a%3Db。


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