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匿名 發問於 教育及參考書小學及中學教育 · 2 年前

求√(x^4-15x^2-8x+80)-√(x^4-3x^2+4)的最大值?

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  • 2 年前
    最愛解答

    sqrt(x^4-15x^2-8x+80)-sqrt(x^4-3x^2+4)=sqrt[(x^2-8)^2+(x-4)^2]-sqrt[(x^2)+x^2] =d(P,A)-d(P,B), where A(4,8), B(0,2), and P(x,x^2) is a point allowed to move on the parabola y=x^2. A rough sketch on xy-plane shows that point A,B are on different sides of the parabola y=x^2. By triangle inequality, 0<d(P,A)-d(P,B)<=d(A,B)=sqrt(52), and the equality holds when A,B,P are collinear: either P-B-A or B-P-A. When P-B-A, d(P,A)-d(P,B)=sqrt(52); when B-P-A, d(P,A)-d(P,B)<sqrt(52). Besides, at infinity (+/-), one can show that limit of {sqrt(x^4-15x^2-8x+80)-sqrt(x^4-3x^2+4)=sqrt[(x^2-8)^2+(x-4)^2]-sqrt[(x^2-2)^2+x^2]}=|(x^2-8)| -|(x^2-2)|, goes to 6<sqrt(52). So the max value is sqrt(52).

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