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Math Help - Differentiation.. the values of x for which dy/dx > 0

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    Differentiation.. the values of x for which dy/dx > 0

    Given that  y=x^2-6x+9 , the values of x for which  \frac {dy}{dx} \geq 0

    This is what I'm thinking.

    First differentiate. Then put the sign then 0. You will get  2x-6 \geq 0

    Is this right?
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  2. #2
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    Quote Originally Posted by jgv115 View Post
    Given that  y=x^2-6x+9 , the values of x for which  \frac {dy}{dx} \geq 0

    This is what I'm thinking.

    First differentiate. Then put the sign then 0. You will get  2x-6 \geq 0

    Is this right?
    Yes,
    from there you find the range of x for which f'(x)\ \ge\ 0

    In the case of quadratics, you can alternatively find the axis of symmetry.
    The co-efficient of x^2 is positive, so the graph is U-shaped.
    Hence the slope of the tangent is \ge\ 0 at or after the minimum,
    which occurs halfway between the roots or at the roots in the case of a double root.

    x^2-6x+9=(x-3)(x-3)

    Double root at x=3, so the tangent has a positive slope after x=3.
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