1、判断函数的奇偶性
(1)解:
f(-x)=(-x)cos(-x)=-xcos(x)=-f(x)
所以,y=xcos(x)是奇函数
(3)解:
f(-x)={e^(-x)-e^(x)}/2=-{e^(x)-e^(-x)}/2=-f(x)
所以,y={e^(x)-e^(-x)}/2是奇函数
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/023b5bb5c9ea15ce1942c291a4003af33b87b2fd?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:f(0)=1-2×0=1,f(-1)=1-2×(-1)=3,f(1)=1-2×1=-1,f(-2)=(-2)²+1=5,f(2)=2²+1=5
3、解:由于3x=y+2,x=y/3+2/3,则函数y=3x-2的反函数为
y=x/3+2/3
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b812c8fcc3cec3fd5e9685edc488d43f869427f9?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)