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匿名 發問於 科學及數學數學 · 6 年 前

More about completing square

(1) If x is real, find the minimum value of x^4 + 6x^2 - 7 by completing square.

(2) Let f(x) = 2x/(x^2+x+1).

If x is real, find the values of x such that f(x) attains its minimum and maximum respectively.

(3) Slove √x+√(y-1)+√(z-2)=(x+y+z)/2

2 個解答

評分
  • 6 年 前
    最佳解答

    1)x⁴+ 6x² - 7

    = x⁴+ 2(3)x² + 9 - 16

    = (x² + 3)² - 16

    ≥ (0 + 3)² - 16

    = - 7

    ∴ The minimum value of x⁴+ 6x² - 7 is 7.

    Alternatively :

    x⁴+ 6x² - 7

    = (x²)² + 6x² - 7 ≥ 0² + 6(0)² - 7 = - 7 .

    2)...... 2x

    ───────

    x² + x + 1........... 2

    = ───────

    ... x + 1 + 1/x

    ............. .......2

    = ──────────────

    ... x + 2√x(1/√x) + 1/x - 1

    .............. 2

    = ──────────

    ... (√x + 1/√x)² - 1

    When (√x + 1/√x)² = 0 , f(x) attains its minimum.

    i.e.

    x + 1/x + 2 = 0

    x² + 2x + 1 = 0

    (x + 1)² = 0

    x = - 1∴ f(x) attains its minimum at x = - 1.

    ...... 2x

    ───────

    x² + x + 1........... 2

    = ───────

    ... x + 1 + 1/x

    ............. .......2

    = ──────────────

    ... x - 2√x(1/√x) + 1/x + 3

    ............. ..2

    = ──────────

    ... (√x - 1/√x)² + 3

    When (√x - 1/√x)² = 0 , f(x) attains its maximum.

    i.e.

    x + 1/x - 2 = 0

    x² - 2x + 1 = 0

    (x - 1)² = 0

    x = 1 ∴ f(x) attains its maximum at x = 1.

    3)√x + √(y - 1) + √(z - 2) = (x + y + z) / 2

    x - 2√x + y - 2√(y - 1) + z - 2√(z - 2) = 0

    (x - 2√x + 1) + (y - 1 - 2√(y - 1) + 1) + (z - 2 - 2√(z - 2) + 1) = 1 - 1 + 1 - 2 + 1

    (√x - 1)² + (√(y - 1) - 1)² + (√(z - 2) - 1)² = 0

    (√x - 1)² = (√(y - 1) - 1)² = (√(z - 2) - 1)² = 0

    √x = 1 and √(y - 1) = 1 and √(z - 2) = 1

    x = 1 and y = 2 and z = 3

  • 6 年 前

    The first two questions are answered here:

    http://imgur.com/qq0Wcro

    For the last one, not much idea.

    Seems that the level is different from the other two?

    2013-10-17 00:59:28 補充:

    我又慢左啦~

    只可以排第三......

    2013-10-17 01:07:06 補充:

    好多謝你的指導呀~

    http://imgur.com/rlrr7hP

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