Maths - Point of inflection, Maximum and minimum points.

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BrandonS03
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So I know that the second detrivative tells us the rate of change of the gradient of a function and tells us the maximum or minimum points by seeing whether its positive or negative for all values of x that are obtained by letting dy/dx =0 but with y=x^3 we have dy/dx = 3x^2 and 3x^2=0 means that x=0 and the second derivative tells us that at x=0 the rate of change of the gradient is 0 so its a point of inflection. Does that mean that y=x^3 has no maximum or minimum points?
Also is the point of inflection known as the point when the concavity of a function changes from either positive to negative or negative to positive?
Also when i put the graphs of y=x^3 6x^2 5 and its derivative y=3x^2 12x on the same axis i found the x coordinate for the minimum point of the derivative is the same as the x coordinate as the point of inflection for the original cubic function. How does this work and whats the link here? What if we had the derivative as a cubic so we had multiple maximums and minimums does that mean we have multiple points of inflection too in the original function?
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RDKGames
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(Original post by BrandonS03)
So I know that the second detrivative tells us the rate of change of the gradient of a function and tells us the maximum or minimum points by seeing whether its positive or negative for all values of x that are obtained by letting dy/dx =0 but with y=x^3 we have dy/dx = 3x^2 and 3x^2=0 means that x=0 and the second derivative tells us that at x=0 the rate of change of the gradient is 0 so its a point of inflection. Does that mean that y=x^3 has no maximum or minimum points?
Yes the function has no local maxima/minima if all the stationary points are points of inflection.

Also is the point of inflection known as the point when the concavity of a function changes from either positive to negative or negative to positive?
It's when \dfrac{d^2 y}{dx^2} > 0 on one side of the point, and \dfrac{d^2 y}{dx^2} < 0 on the other side.

Also when i put the graphs of y=x^3 6x^2 5 and its derivative y=3x^2 12x on the same axis i found the x coordinate for the minimum point of the derivative is the same as the x coordinate as the point of inflection for the original cubic function. How does this work and whats the link here?
Well think about what a point of inflection means. It means determining where \dfrac{d^2 y}{dx^2} = 0 first of all.

If I denote the gradient function as Y = \dfrac{dy}{dx}, then the above condition is the same as saying \dfrac{dY}{dx} = 0. And this equation gives precisely the stationary points of Y.
This means that points of inflection of f(x) must be where f'(x) obtains its local maxima/minima points.

Since \dfrac{d}{dx}(3x^2+12x) = 0 at x=-2, then the (possible) point of inflection must be at x=-2 on the curve.

What if we had the derivative as a cubic so we had multiple maximums and minimums does that mean we have multiple points of inflection too in the original function?
Again, it doesn't imply we have multipe points of inflection, it just implies we have multiple candidates to be our points of inflection. Remember that, in addition to f''(x)=0, we must also have that f'''(x) \neq 0 to class these points as inflection points.

But indeed we can have multiple points of inflection. Take the function y=\dfrac{1}{4}x^4 - \dfrac{9}{2}x^2.

Here the possible points of inflection are found by setting \dfrac{d^2y}{dx^2} = 3x^2-9 = 0 \implies x = \pm \sqrt{3}.
The third derivative, \dfrac{d^3 y}{dx^3} = 6x \neq 0 when x=\pm \sqrt{3} hence we have two points of inflection.

For a graph like y = \sin x we have infinitely many inflection points.
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BrandonS03
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(Original post by RDKGames)
Again, it doesn't imply we have multipe points of inflection, it just implies we have multiple candidates to be our points of inflection. Remember that, in addition to f''(x)=0, we must also have that f'''(x) \neq 0 to class these points as inflection points.

But indeed we can have multiple points of inflection. Take the function y=\dfrac{1}{4}x^4 - \dfrac{9}{2}x^2.

Here the possible points of inflection are found by setting \dfrac{d^2y}{dx^2} = 3x^2-9 = 0 \implies x = \pm \sqrt{3}.
The third derivative, \dfrac{d^3 y}{dx^3} = 6x \neq 0 when x=\pm \sqrt{3} hence we have two points of inflection.

For a graph like y = \sin x we have infinitely many inflection points.
Regarding your last statement y=sin(x) has infinitrly many inflection points. How would you go to prove this? Is it more advanced or something else?
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RDKGames
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(Original post by BrandonS03)
Regarding your last statement y=sin(x) has infinitrly many inflection points. How would you go to prove this? Is it more advanced or something else?
You would find where \dfrac{d^2 y}{dx^2} = 0, i.e. you need to determine where -\sin x = 0. (Solution: x=\pi k for k \in \mathbb{Z})

Then you test these in the third derivative, \dfrac{d^3 y}{dx^3} = -\cos x. And -\cos(\pi k) = (-1)^{k+1} so it's never 0. Hence all these points x=\pi k are inflection points - there are infinitely many of them.
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BrandonS03
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(Original post by RDKGames)
You would find where \dfrac{d^2 y}{dx^2} = 0, i.e. you need to determine where -\sin x = 0. (Solution: x=\pi k for k \in \mathbb{Z})

Then you test these in the third derivative, \dfrac{d^3 y}{dx^3} = -\cos x. And -\cos(\pi k) = (-1)^{k+1} so it's never 0. Hence all these points x=\pi k are inflection points - there are infinitely many of them.
How does the third derivative become involved?
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RDKGames
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(Original post by BrandonS03)
How does the third derivative become involved?
Like I said, for a point of inflection you need that \dfrac{d^2 y}{dx^2} < 0 on one side of the point and \dfrac{d^2 y}{dx^2} > 0 on the other.

So this means our function \dfrac{d^2 y}{dx^2} must be either strictly increasing as we pass through our point, or strictly decreasing.

Hence its derivative, ie  \dfrac{d^3 y}{dx^3} , must be either >0 on the point or <0 on the point. Or, to put it equivalently,  \dfrac{d^3 y}{dx^3} \neq 0
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