有关函数单调性&奇偶性
已知f(x)为奇函数,且f(x+2)=-f(x),当0≤x≤1,f(x)=x.求f(7.5)的值.
谢谢解答~
参考答案:because f(x+2)=-f(x):
f(7.5)=f(5.5+2)=-f(5.5)=-f(3.5+2)=-(-f(3.5))=f(3.5)=f(1.5+2)=-f(1.5)=-f(-0.5+2)=-(-f(-0.5))=f(-0.5)
And f(x)=-f(-x) （f(x)奇函数）
f(7.5)=f(-0.5)=-f(0.5)
又当0≤x≤1,f(x)=x
f(7.5)=-f(0.5)=-0.5