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.
Best Answer
Check you zero's again: $x_1 = -1$ is a zero, as is $x_2 = -5/3$.
Try evaluating $f(x)$ at $x = -3/2$, i.e., $x = -2$ and compare with find $f(-5/3)$
Likewise for $f(-1)$. Choose smaller intervals around each critical point. Try, say, evaluating $f(x)$ at $x = -3/2$ and $x = -1/2$, to compare with $f(-1)$.
Since you have two critical points with only $2/3$ of a unit separating them, you need smaller intervals to determine the behavior of the function near those point.
You can also use the sign of the derivative to determine on which interval(s) a function is increasing, and when it is decreasing. When $f'(x) > 0 \implies f(x)$ is increasing, when $f'(x) \lt 0 \implies f(x) $ is decreasing. But again, you'll want to evaluate $f'(x)$ for $x$ very near the critical points $x_1, x_2$