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Math Help - Derivatives

  1. #1
    Member Mr Rayon's Avatar
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    Derivatives

    Given that y=2x^3-6x^2+18x, find \frac{dy}{dx} Hence show that \frac{dy}{dx}> 0 for all x.

    I know how to find the derivative but what does it mean by:

    Hence show that \frac{dy}{dx}> 0 for all x, above? By finding the derivative how do I show this? If the derivative is positive do I assume its greater than 0? Is that how I show it?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Mr Rayon View Post
    Given that y=2x^3-6x^2+18x, find \frac{dy}{dx} Hence show that \frac{dy}{dx}> 0 for all x.

    I know how to find the derivative but what does it mean by:

    Hence show that \frac{dy}{dx}> 0 for all x, above? By finding the derivative how do I show this? If the derivative is positive do I assume its greater than 0? Is that how I show it?
    y=2x^3-6x^2+18x

    \frac{dy}{dx}=6x^2-12x+18

    \frac{dy}{dx}=6(x^2-2x+3)

    \frac{dy}{dx}=6[(x-1)^2+2]

    Hence \frac{dy}{dx}>0 for all x.
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