已知函数f(x)=1/3 x^3+mx^2-3m^2x+1(m>0)(2)若函数f(x)在区间(2m-1,m+1)上单调递增,求实数m的取值

问题描述:

已知函数f(x)=1/3 x^3+mx^2-3m^2x+1(m>0)(2)若函数f(x)在区间(2m-1,m+1)上单调递增,求实数m的取值

f'(x)=x²+2mx-3m²=(x+3m)(x-m)∵m>0∴f'(x)>0即(x+3m)(x-m)>0解得xm∴f(x)的单调递增区间为(-∞,-3m)(m,+∞)∵函数f(x)在区间(2m-1,m+1)上单调递增∴m+1≤-3m且m+1>2m-1解得m≤-1/4且m0)或2m-1≥m且m...