设f(x)=
{x3+2x2+2,x≤0,
{1-x2,x>0,
求f(x+1).
【正确答案】:f(x+1)= {(x+1)3+2(x+1)2+2,x+1≤0, {1-(x+1)2, x+1>0, 化简得f(x+1)= {x3+5x2+7x+5,x≤-1, { -x2-2x, x>-1.
设f(x)={x3+2x2+2,x≤0,{1-x2,x>0,求f(x+1).
- 2024-11-07 09:15:32
- 高等数学(经管类)(13125)