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Math Help - Calculus question!

  1. #1
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    Calculus question!

    'The gradient of a curve is given by (dy)/(dx) = x^2 - 6x. Find the set of values of x for which y is an increasing function of x.

    Im pretty good at calculus, but i'm not sure what it means by an increasing function of x.

    Any help?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nugiboy View Post
    'The gradient of a curve is given by (dy)/(dx) = x^2 - 6x. Find the set of values of x for which y is an increasing function of x.

    Im pretty good at calculus, but i'm not sure what it means by an increasing function of x.

    Any help?
    What does the derivative of a function mean? Thus if it is positive/negative what does that mean about the function?

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    What does the derivative of a function mean? Thus if it is positive/negative what does that mean about the function?

    -Dan
    Ah right so i need to find the values of x for which the gradient is positive?
    Do i just do this by trial and error substituting in x values, or is there something im missing?
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  4. #4
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    Quote Originally Posted by nugiboy View Post
    Ah right so i need to find the values of x for which the gradient is positive?
    Do i just do this by trial and error substituting in x values, or is there something im missing?
    You want x^2 - 6x > 0

    If x < 0, then x(x-6) > 0
    If 0 < x < 6, then x(x-6) < 0
    If x > 6, then x(x-6) > 0

    So if x \in ( - \infty, 0) \bigcup (6,\infty), then \frac{dy}{dx} is positive.

    Or in English, except for values of x between 0 and 6, everywhere it is increasing
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  5. #5
    Moo
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    Hello,

    A table of signs can help :

    x \leq 0 \text{ iff } x \leq 0 (this can look unuseful, but it's the reasoning )

    x-6 \leq 0 \text{ iff } x \leq 6

    So if x \in ]-\infty,0] \ , \ x \leq 0 \text{ and } x-6 \leq 0. \text{ Hence } x(x-6) \geq 0

    If x \in [0,6] \ , \ x \geq 0 \text{ and } x-6 \leq 0. \text{ Hence } x(x-6) \leq 0

    If x \in [6,+\infty[ \ , \ x \geq 0 \text{ and } x-6 \geq 0. \text{ Hence } x(x-6) \geq 0


    So the function will be increasing for \boxed{x \in ]- \infty,0] \cup [6,+\infty[} \ (= \mathbb{R} \backslash [0,6])
    Last edited by Moo; April 24th 2008 at 04:26 AM.
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