Can someone explain the link between the turning point (local max, min & stationary point of inflection) and it's relationship to derivatives.
Let me clarify what I understand (feel free to correct me).
If we derive an equation and let it equal $0$, the value of $x$ is some kind of turning point.
To find out what kind, we derive again, and sub. that value of $x$ , and look for the following.
If $x > 0$ it is a local min
If $x < 0$ is it a local max
If $x = 0$ it is an inflection point.
For example
$$y= (x-1)^3 + 1$$
Is this a local max/min/inflection point at $(1,\space 1)$
So derive $1$st time $= 3(x-1)^2 = 3x^2 -6x +3$
we let it equal $0$ so
$0 = 3x^2 -6x +3$
$x =1$
Then we derive again to know what kind of turning point.
Derive $2$nd time $= 6x -6$
Sub. $x=1$
$6(1)-6 = 0$
Therefore it is an inflection point??
Is this right… hmmmm.
Best Answer
The derivative provides information about the gradient or slope of the curve/graph of a function which can be used to locate points on the function's curve/graph where its gradient is $0$. These points are often associated with the largest or smallest values of the functions.
A stationary point is any point where the tangent to the curve/graph is horizontal. These can be calculated by finding the derivative and equating it to $0$. i.e. stationary points can be located by looking for points at which $\frac{dy}{dx}=0$.
At a turning point, $\frac{dy}{dx}=0$. So all turning points are stationary points. But not all points where $\frac{dy}{dx}=0$ are turning points, i.e. not all stationary points are turning points.
The local maximum can be defined as a point where $f(a)\ge f(x)$ for all $x$ in the interval in layman's terms $f(a)$ is the highest point in the interval. While the local minimum is the opposite, $f(a)\le f(x)$ for all $x$ in the interval aka $f(a)$ is the lowest point in the interval.
Now, the link between all these terms and the derivative can be discovered when you examine the gradient of the curve/graph. Think about the graph of $y=x^2$ (you can just envision a parabola). Clearly at the bottom of the curve $\frac{dy}{dx}=0$, from the left $\frac{dy}{dx}$ is negative (because the slope is going down) and from the right $\frac{dy}{dx}$ is positive (because the slope is going up). $\frac{dy}{dx}$ goes from negative to $0$, to positive and thus is clearly increasing.
The second derivative is used to verify that a point is a local maximum or minimum. If the second derivative is positive then $\frac{dy}{dx}$ would be increasing and that stationary point would be a minimum turning point.
In conclusion, if $\frac{dy}{dx}=0$ (is a stationary point) and if $\frac{d^2y}{dx^2}>0$ at that same point, them the point must be a minimum. This can also be observed for a maximum turning point. If $\frac{dy}{dx}=0$ (is a stationary point) and if $\frac{d^2y}{dx^2}<0$ at that same point, them the point must be a maximum. You can read more here for more in-depth details as I couldn't write everything, but I tried to summarize the important pieces.
A point of inflection on the graph/curve is a point where the gradient stops increasing and starts decreasing, or stops decreasing and starts increasing. What you have is correct, since $\frac{d^2y}{dx^2}=0$, at $(1,\space 1)$ that is indeed an inflection point. As Fly by Night suggested, try youtubing videos, Khan Academy is a good source.