题目

已知函数f(x)=kx^3-3(k+1)x^2-k^2+1(k>0),函数f(x)的单调减区间为(0,4),当x>k,求证2√x>3-1/x

提问时间：2020-10-11

答案

f'(x)=3kx^-6(k+1)x=0,x1=0,x2=2(k+1)/k,k>0,f(x)在（x1,x2)上递减,其余情况递增,已知f(x)的单调减区间为(0,4),∴2(k+1)/k=4,k+1=2k,k=1.f(x)=x^3-6x^2,x>k=1时2√x>3-1/x,①平方得4x>9-6/x+1/x^2,4x^3>9x^2-6x+1,4x...