Question

设f(x)=x³-2x²+3x+3,求f(0),f(1),f(-1),f(x+1)

Answer 1

解：f（0）= 0 ³ - 2 × 0 ² + 3 × 0 + 3 = 3
f（1）= 1 ³ - 2 × 1 ² + 3 × 1 + 3 = 5
f（- 1）= （- 1）³ - 2 × （- 1）² + 3 × （- 1）+ 3 = - 3
f（x + 1）= （x + 1）³ - 2（x + 1）² + 3（x + 1）+ 3
= （x + 1）（x ² + 2 x + 1）- 2（x ² + 2 x + 1）+ 3 x + 3 + 3
= x ³ + x ² + 2 x ² + 2 x + x + 1 - 2 x ² - 4 x - 2 + 3 x + 6
= x ³ + x ² + x + 5

Answer 2

解
f(0)=0³-2×0²+3×0+3=3
f(-1)=(-1)³-2×(-1)²+3×(-1)+3
=-1-2-3+3
=-3

f(1)=1³-2×1²+3×1+3
=1-2+3+3
=5

f(x+1)=(x+1)³-2(x+1)²+3(x+1)+3
=x³+3x²+3x+1-2(x²+2x+1)+3x+6
=x³+x²+2x+5