f'(x)=5x^4+20x^3+15x^2
=5x^2(x^2+4x+3)
=5x^2(x+1)(x+3)
則f'(x)≥0的解為x≤-3或x≥-1；
f'(x)≤0的解為-3≤x≤-1或x=0.
所以,
f(x)在(-∞,-3]遞增,在[-3,-1]遞減,在[-1,＋∞)遞增
所以,在[-1,4]上,f(x)是遞增的,
最小值為f(-1)=-1+5-5+1=0.