x^4-4x^3+10x^2-12x+9
=(x^4-4x^3+6x^2-4x+1)+(4x^2-8x+4)+4
=[(x^4-4x^3+4x^2)+(2x^2-4x)+1]+4(x^2-2x+1)+4
=[(x^2-2x)^2+2(x^2-2x)+1]+4(x^2-2x+1)+4
=[(x^2-2x)+1]^2+4(x^2-2x+1)+4
=[(x-1)^2]^2+4(x^2-2x+1)+4
=(x-1)^4+4(x-1)^2+4
其中：(x-1)^4≥0、4(x-1)^2≥0,當(x-1)^4=0、4(x-1)^2=0時,即x=1,原式獲得最小值為4.
所以
x^4-4x^3+10x^2-12x+9的最小值是4.