解由f(x)=x3-3x2-9x
得f'(x)=3x^2-6x-9
令f'(x)＞0
即3x^2-6x-9＞0
即x^2-2x-3＞0
即（x-3)(x+1)＞0
即x＞3或x＜-1
故f(x)的单调递增区间为（负无穷大,-1）和（3,正无穷大）.