全文预览

复旦陈传璋《数学分析》答案.pdf

上传者:qnrdwb |  格式:pdf  |  页数:285 |  大小:0KB

文档介绍
?(x 2) = (x 2) 3+1 =x 6+1,(?(x)) 2= (x 3+1) 2=x 6+2x 3+1, ?(?(x)) = (x 3+1) 3+1 =x 9+3x 6+3x 3+2 10 6.?f(x) = 1 1?x §|f(f(x)), f(f(f(x))), f ?1 f(x) ?. )μ?f(x) = 1 1?x §K f(f(x)) = 1 1? 1 1?x = x?1 x , f(f(f(x))) = 1 1? 1 1? 1 1?x = 1 1? x?1 x =x, f ?1 f(x) ?= 1 1?(1?x) = 1x 7.|e??ê???ê9??ê????μ (1)y=x 2(?∞< x60) (2)y= √1?x 2(?16x60) (3)y= sinx ?π2 6x6 32 π?(4)y= ??? x, ??∞< x <1? x 2, ?16x64? 2 x, ?4< x <+∞? )μ (1)?y=x 2(?∞< x60)§Kx=?√y(06y <+∞)§l d?ê???ê?y=?√x(06y <+∞) (2)?y= √1?x 2(?16x60)§Kx=? p1?y 2(06y61)§l d?ê???ê?y=?√1?x 2(06 x61) (3)?y= sinx ?π2 6x6 32 π?§Kx=π?arcsiny(?16y61)§l d?ê???ê?y=π? arcsinx(?16x61) (4)?y= ??? x, ??∞< x <1? x 2, ?16x64? 2 x, ?4< x <+∞? §Kx= ??? y, ??∞< y <1? √y, ?16y616? log 2y, ?16< x <+∞? §l d?ê???ê ?y= ??? x, ??∞< x <1? √x, ?16x616? log 2x, ?16< x <+∞? .

收藏

分享

举报
下载此文档