f(x-1)=x^2+2x-1,求f(x)

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f(x-1)=x^2+2x-1,求f(x)f(x-1)=x^2+2x-1,求f(x)f(x-1)=x^2+2x-1,求f(x)设a=x-1,则x=a+1f(x-1)=f(a)=x^2+2x-1=(a+

f(x-1)=x^2+2x-1,求f(x)
f(x-1)=x^2+2x-1,求f(x)

f(x-1)=x^2+2x-1,求f(x)
设 a = x - 1,则 x = a + 1
f(x-1)= f(a) = x^2+2x-1 = (a+1)^2 + 2(a+1) -1 = a^2 + 4a + 2
所以,f(x) = x^2 + 4x + 2

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

设 t = x - 1,则 x = t + 1,代入左式:
f(t + 1 - 1) = (t + 1)^2 + 2(t + 1) - 1
即可求出 f(t) = 多少 t,
然后再把 t 换成 x 就OK了。

令x=x+1,则f(x+1-1)=(x+1)^2+2(x+1)-1,
所以f(x)=x^2+4x+2

f(x-1)=x²+2x-1
令x-1=t
x=t-1
f(t)=(t-1)²+2(t-1)-1
=t²-2
f(t-1)=(t-1)²-2
=t²-2t-z
x=t-1
f(x)=t²-2t-1