设函数f(x)=-1/3x^3+x^2+(m^2-1)x (x∈R),其中m>0,求函数f(x)的单调区间和极值.

f(x)=-(1/3)x^3+x^2+(m^2-1)x对x求导,得f'(x)=-x^2+2x+(m^2-1)=-(x-1)^2+m^2=-(x+m-1)(x-m-1)当f'(x)＞0时,[x-(1-m)][x-(m+1)]＜0∵m＞0,∴1+m＞1-m∴1-m＜x＜1+m即在区间(1-m,1+m)内原来的函数单调递增f'(x)≤0时,[...