Question

判斷下列函數(shù)的奇偶性.

(1)  (a＞0,且a≠1);

(2);

(3);

(4)

Answer 1

解析:(1)由 ＞0,得-1＜x＜1.?

∴定義域關(guān)于原點(diǎn)對(duì)稱.?

又∵f(x)+f(-x)=,?

∴f(-x)=-f(x).?

∴f(x)是奇函數(shù).?

(2)由F(x)=x(+)易得其定義域?yàn)?I >x∈ (-∞,0)∪(0,+∞),關(guān)于原點(diǎn)對(duì)稱.?

∵F(-x)=-x(+)=x(-)?

=x(-)=x(+)=F(x),?

∴F(x)是偶函數(shù).?

(3)非奇非偶函數(shù).因?yàn)?I >x=-時(shí),f(x)無(wú)意義,但x=時(shí),f(x)有意義,故f(x)的定義域關(guān)于原點(diǎn)不對(duì)稱.?

(4)因?yàn)?I >f(x)的定義域?yàn)镽,?

當(dāng)x＞0時(shí),-x＜0,f(-x)=(-x)²-1=x²-1=-f(x);??

當(dāng)x＜0時(shí),-x＞0,f(-x)=1-(-x)²=1-x²=-f(x).??

但當(dāng)x=0時(shí),f(-x)=1=f(x),故f(x)是非奇非偶函數(shù).