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Math Help - Differentiation Question...

  1. #1
    Member looi76's Avatar
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    Differentiation Question...

    Question:
    For the graph x^2-8x+4 :
    (i) find the coordinates of the stationary point;
    (ii) say, with reasoning, weather this is a maximum or minimum point;
    (iii) check your answer by using the method of 'completing the square' to find the vertex;
    (iv) state the range of values which the function can take.

    Attempt:
    (i) f(x) = x^2-8x+4
    f'(x) = 2x-8
    f'(x) = 0
    2x-8 = 0
    2x = 8
    x = \frac{8}{2} = 4

    f(x) = x^2-8x+4
    f(x) = 4^2-8\times4+4
    f(x) = -12

    Stationary Point = (4,-12)

    (ii) f'(1) = 2x-8
    f'(1) = 2(1)-8
    f'(1) = -6

    f'(7) = 2x-8
    f'(7) = 2(7)-8
    f'(7) = 6

    x \leq 4
    Minimum Point

    (iii) Need help...
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:
    For the graph x^2-8x+4 :
    (i) find the coordinates of the stationary point;
    (ii) say, with reasoning, weather this is a maximum or minimum point;
    (iii) check your answer by using the method of 'completing the square' to find the vertex;
    (iv) state the range of values which the function can take.

    Attempt:
    (i) f(x) = x^2-8x+4
    f'(x) = 2x-8
    f'(x) = 0
    2x-8 = 0
    2x = 8
    x = \frac{8}{2} = 4

    f(x) = x^2-8x+4
    f(x) = 4^2-8\times4+4
    f(x) = -12

    Stationary Point = (4,-12)

    (ii) f'(1) = 2x-8
    f'(1) = 2(1)-8
    f'(1) = -6

    f'(7) = 2x-8
    f'(7) = 2(7)-8
    f'(7) = 6

    x \leq 4
    Minimum Point

    (iii) Need help...
    x^2-8x+4 = (x^2 - 8x + 16) - 16 + 4 = (x - 4)^2 - 16 + 4 = (x - 4 )^2 - 12. The quadratic is now in turning point form. By inspection, the turning point is at (4, -12).
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