Question

已知函数f(x)=x^n-x^-n\x^n+x^-n(x>0,n属于N正 判断f(x)在(0,正无穷)上的单调性

Answer 1

f(x)=(x^n-x^-n)/(x^n+x^-n)

=[x^(2n)-1]/[x^(2n)+1]

=1-2/[x^(2n)+1]

因x>0,n∈N+,故随着x的增大,x^(2n)+1增大,2/[x^(2n)+1]减小,1-2/[x^(2n)+1]增大.故

f(x)在(0,+∞)上的单调递增.