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Differentiation problem!

  1. Feb 2, 2008 #1
    Question:
    For the graph [TEX]x^2-8x+4 [/TEX] :
    (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) [TEX]f(x) = x^2-8x+4[/TEX]
    [TEX]f'(x) = 2x-8[/TEX]
    [TEX]f'(x) = 0[/TEX]
    [TEX]2x-8 = 0[/TEX]
    [TEX]2x = 8[/TEX]
    [TEX]x = \frac{8}{2} = 4[/TEX]

    [TEX]f(x) = x^2-8x+4[/TEX]
    [TEX]f(x) = 4^2-8\times4+4[/TEX]
    [TEX]f(x) = -12[/TEX]

    Stationary Point = [TEX](4,-12)[/TEX]

    (ii) [TEX]f'(1) = 2x-8[/TEX]
    [TEX]f'(1) = 2(1)-8[/TEX]
    [TEX]f'(1) = -6[/TEX]

    [TEX]f'(7) = 2x-8[/TEX]
    [TEX]f'(7) = 2(7)-8[/TEX]
    [TEX]f'(7) = 6[/TEX]

    [TEX]x \leq 4[/TEX]
    Minimum Point

    (iii) Need help...
     
  2. jcsd
  3. Feb 3, 2008 #2
    Here is not the place for homework questions...

    (read the first thread)
     
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