Question

幂函数过点p（1/2,4）,f(x)=x^-2

Answer 1

解由f(x)=x^(-2)=1/x^2则知1/x^2＞0即f(x)＞0故函数的值域为（0,正无穷大）又由f(-x)=1/(-x)^2=1/x^2=f(x)故f(x)是偶函数(2)设x1,x2属于(0,+∞)且x1＜x2则f(x1)-f(x2)=1/x1^2-1/x2^2=(x2^2-x1^2)/x1^2x2^2由0＜x1＜x...