ruize
又g(-2) =f(-2) + 8 = 18,所以f(2) = g(2)-8 =-26. (5) 解:?x???1,0???0,1?;??x???1,0???0,1?;
1?x21?x2??0 ?f(x)?f(?x)?xx1?x2所以f?x??是奇函数。
x?2?2课后作业
(1) C提示:由于f(?x)=-f(x),所以f(x)·f(?x)=-[f(x)]≤0,故选C. (2) B 提示:f?7.5???f?5.5??f?3.5???f?1.5??f??0.5???f?0.5???0.5. (3)C 提示:A、B、D所给函数的定义域都不关于原点对称.
(4) D解析: 令F(x)?f(x)?4?ax?bx,则F(x)?ax?bx为奇函数 F(?2)?f(?2)?4?6,F(2)?f(2)?4??6,f(2)??10 (5) 0提示:∵f?x?是偶函数,∴f?331???f[?(2?1)]?f?1?2???2?1,
?∴f1?2?f???1???0.
?1?2??(6)解析:在f?ab??af?b??bf?a?中,取a?b?1,得f?1??2f?1?,∴f?1??0, 又∵f?1???f??1??f??1?,∴f??1??0, 在f?ab??af?b??bf?a?中,
取a??1,b?x,有f??x??f??1?x???f?x??xf??1???f?x?, 即f(?x)??f(x),故f?x?为奇函数.
1.3第3课时 函数的奇偶性
ruize又g(-2)=f(-2)+8=18,所以f(2)=g(2)-8=-26.(5)解:?x???1,0???0,1?;??x???1,0???0,1?;1?x21?x2??0?f(x)?f(?x)?xx1?x2所以f?x??是奇函数。x?2?2课后作业(1)C提示:由于f(?x)=-f(x),所以
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