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.
$\endgroup$3 Answers
$\begingroup$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 $x_1, x_2$
$\endgroup$ 13 $\begingroup$Picking nearby points to see if a critical point is a maximum or minimum can be hazardous because the curve might do things you don't expect. This cubic is very close to flat between the zeros of the derivative. The reason $f(-2)=f(-1)=4$ is that you have gone over the hill and started down the other side. You shouldn't pick test points that are beyond another critical point. If you had tried $f(-\frac 32)=\frac {33}8$ and $f(-\frac 12) =\frac {35}8$ you would see that $-1$ is a minimum.
You could also try the second derivative test to find whether these are maxima or minima.
$\endgroup$ 2 $\begingroup$If you graph the given function and particularly look at the behavior of this cubic (with the emphasis on CUBIC), you can tell with the two stationary points you calculated which one is a max and which is min. You can also look if you want at the derivative, which is quadratic and look at the derivative's graph. When is it above the x-axis and when below? That tells you which stationary point is a max and which is a min. Now you try to find out which is which :)
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