Question

已知函数f(x)满足f（x +1）=x2 +x +1,求fx的解析式

Answer 1

f(x+1)=x²+x+1

设x+1=t,则x=t-1

∴f(t)=f(x+1)=x²+x+1

=(t-1)²+(t-1)+1

=t²-t+1

将t换成x

得到f(x)=x²-x+1

配方：f（x)=(x-1/2)²+3/4

∵x∈[0,2]

∴f(x)min=f(1/2)=3/4

f(x)max=f(2)=3