# [SOLVED] Need help understanding points of inflextion.

• Sep 11th 2009, 01:50 AM
el123
[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?
• Sep 11th 2009, 02:21 AM
Hello el123
Quote:

Originally Posted by el123
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).