數學代數法f(x)

設f(x)=2x²+1. 求f(3x+1)-f(x)

= 2(3x+1)²+1 - ( 2x²+1 )
= 2(9x²+6x+1) - 2x² - 1
=18x² + 12x + 2 - 2x² - 1
=16x² +12x + 1

設f(x)=-x²+x+2.求f(x+1)+f(x-1)

= -(x+1)²+(x+1)+2 + [ -(x-1)²+(x-1)+2 ]
= -(x² + 2x + 1) + 2 + [ -(x² - 2x + 1) + 2 ]
= -x² - 2x - 1 + 2 + [ -x² + 2x - 1 + 2 ]
= -x² - 2x + 1 -x² + 2x + 1
= -2x² + 2

若f(x-1)=2x²-4x,求f(x)

f(x-1)=2x²-4x
= 2x² - 4x + 2 - 2
= 2(x² - 2x + 1) - 2
= 2(x-1)x² - 2

==> f(x) = 2x² - 2



2014-03-04 23:18:21 補充：
設f(x)=x²-5.若f(k)=k+1,求k的值

f(k) = k² - 5 = k + 1

==> k² - k - 6 = 0

(k-3)(k+2) = 0

k = 3 or k = -2

2014-03-04 23:30:17 補充：
若f(x-1)=2x²-4x,求f(x)

f(x-1)=2x²-4x
= 2x² - 4x + 2 - 2
= 2(x² - 2x + 1) - 2
= 2(x-1)² - 2　　　　　　上面打錯了 : = 2(x-1)x² - 2

==> f(x) = 2x² - 2