My question is related to how to find a local maximum and local minimum.
As far I know, for the first we should find derivative of the function and set it to zero. For exmaple, suppose our function is given by:
$$f(x)=x^3+4x^2+5x+6$$
We first differentiate it:
$$f'(x)=3x^2+8x+5$$
For optimal points of $3x^2+8x+5=0$ we find $x_1=-1$ and $x_2=-5/3$.
For local maximum and/or local minimum, we should choose neighbor points of critical points, for $x_1=-1$, we choose two points, $-2$ and $-0$, and after we insert into first equation:
$$f(-2)=4$$
$$f(-1)=-8+16-10+6=4$$
$$f(0)=6$$
So, it means that points $x_1=-1$ is local minimum for this case, right? Because it has minimum output among $-2$ and $-0$, right?
For this case, $f(-2)=f(-1)$, but does it change something? Just consider for first point, so if $f(-1)<f(-2)$, then it means that it would be local minim as well, but if $f(-2)>(-1)$, then it would be saddle point.