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

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

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


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