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Thread: [SOLVED] Need help understanding points of inflextion.

  1. #1
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    [SOLVED] Need help understanding points of inflextion.

    Ok so i am a bit confused with this idea.

    So you take second derivative and set to zero.

    I have this equation

    $\displaystyle f''(x) = 24x + \frac{2}{x^{\frac{3}{2}}} = 0 $

    So i set to zero and get $\displaystyle x = \sqrt[\frac{5}{2}]{\frac{2}{24}} $

    So i take the answer this gives me and plug into f''(x) but i get 17ish. Does this mean i don't have a point on inflextion?

    But if it equaled zero then i would?
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  2. #2
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    Hello el123
    Quote Originally Posted by el123 View Post
    Ok so i am a bit confused with this idea.

    So you take second derivative and set to zero.

    I have this equation

    $\displaystyle f''(x) = 24x + \frac{2}{x^{\frac{3}{2}}} = 0 $

    So i set to zero and get $\displaystyle x = \sqrt[\frac{5}{2}]{\frac{2}{24}} $

    So i take the answer this gives me and plug into f''(x) but i get 17ish. Does this mean i don't have a point on inflextion?

    But if it equaled zero then i would?
    You have sign wrong - it should be $\displaystyle x = \sqrt[\frac{5}{2}]{-\frac{2}{24}} $ - so this equation doesn't have any real roots. And so there's no point of inflexion.

    Yes, there's a point of inflexion if $\displaystyle \frac{d^2y}{dx^2}=0$.

    Bear in mind that $\displaystyle \frac{d^2y}{dx^2}$ gives the rate of change (with respect to x) of $\displaystyle \frac{dy}{dx}$, the gradient of $\displaystyle y = f(x)$.

    So when $\displaystyle \frac{d^2y}{dx^2}>0$, $\displaystyle \frac{dy}{dx}$, the gradient of the curve, is increasing; and when $\displaystyle \frac{d^2y}{dx^2}<0$, $\displaystyle \frac{dy}{dx}$ is decreasing.

    But when $\displaystyle \frac{d^2y}{dx^2}=0$ the gradient of $\displaystyle y = f(x)$ goes from increasing to decreasing (or the other way round). This is what a point of inflexion is - where the tangent touches the curve and cuts it at the same time (in other words has 3-point contact).

    Grandad
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  3. #3
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    Oh yup i see now , thanks mate!
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