√(x＾4-3x^2-6x+13)-√(x^4-x^2+1)的极大值

Question

√(x＾4-3x^2-6x+13)-√(x^4-x^2+1)的极大值

Answer 1

令1/t=√(x＾4-3x^2-6x+13)-√(x^4-x^2+1),则
t=1/[√(x＾4-3x^2-6x+13)-√(x^4-x^2+1)]
=[√(x＾4-3x^2-6x+13)+√(x^4-x^2+1)]/[x^4-3x^2-6x+13-x^4+x^2-1]
=[√(x＾4-3x^2-6x+13)+√(x^4-x^2+1)]/[-2x^2-6x+12]
=-[√(x＾4-3x^2-6x+13)+√(x^4-x^2+1)]/[2(x^2+3x-6)]
当t最小时,1/t最大.
因为t的等号后有个负号,所以分母x^2+3x-6=0时,分母为0,t最小,1/t=原式最大.
解得x=(-3±√33)/2时原式最大
用x^2+3x-6=0化简原式,
原式=√(85-60x)-√(85-60x)=0
所以√(x＾4-3x^2-6x+13)-√(x^4-x^2+1)的极大值是0