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Stationary points of differentiable function `f` are points `x` such that `f'(x)=0.`
Stationary points are often used when searching for local extrema (minimum or maximum) of a function. They are potential extrema of a function, however there are stationary points which are neither maximum nor minimum one such example is
`f'(0)=0` but zero is not an extremum which you can see on the graph below.
So to determine weather a stationary point is minimum or maximum we can use second derivative of a function. The next theorem explains how to do that.
Let `c` be stationary point of twice differentiable function `f.` If `f''(c)<0,` then `f` has local maximum in `c.` If `f''(c)>0,` then `f` has local minimum in `c.`
This is because `f''(c)<0` means that function is decreasing at point `c` and since we know that `f'(c)=0` it follows function is increasing before `c` and decreasing after `c` implying that the point `c` is maximum.
You can also look at it from the point of convex and concave parts of a function which is also tied to the second derivative.
For more on this see the link below.
The second derivative of a function determines its concavity. If the second derivative at a point is positive and therefore concave upward, then that means the function is "smiling" (u-shape) at that point. On the other hand, if the second derivative at a point is negative and therefore concave downward, then the function is "frowning" (upside-down u-shape) at that point. When the second derivative at a point evaluates to 0, that means there is an inflection point there. Intuitively, if a positive second derivative means the graph is smiling there and a negative second derivative means the graph is frowning there, then when the second derivative is equal to 0 it must be in some sort of in-between stage, which is exactly what an inflection point - a point when the graph changes between being concave up and being concave down (doesn't matter which way it is changing). Visually, an exactly of an inflection point is the point (0, 0) on the graph of y = x^3.
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