Calculus

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The graph of f(x)=(ax+b)/(x^2 - 5x + 4) has a horizontal tangent line at (2, -1). Find a and b

  • Calculus -

    f'(x) = ( a(x^2 - 5x + 4) - (2x-5)(ax+b) )/(x^2 - 5x +4)^2
    = 0 for a horizontal line

    when x = 2
    a(-2) + (2a+b) = 0
    -2a + 2a + b = 0
    b = 0

    also f(2) = -1
    2a/(4 - 10 + 4) = -1
    2a/-2 = 1
    a = 1

    a=1 , b=0

    so f(x) = x/(x^2 - 5x + 4)

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