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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:


We first differentiate it:


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:




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.

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up vote 3 down vote accepted

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 $x_1, x_2$

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but actually if at critical point function is maximum among the two it's neigbour points,then this point is called local local maximum right? – dato datuashvili Mar 27 '13 at 19:27
i have update something (fix up errors in my question ) please see it – dato datuashvili Mar 27 '13 at 19:28
Depends on how far apart your neighbors are. After all one could take a sine wave function and change it's period so that for each integer value the function was a local maximum so thus you have to be careful about how far apart you are looking at things. – JB King Mar 27 '13 at 19:29
You still need smaller intervals (neighbors closer to the critical points). – amWhy Mar 27 '13 at 19:32
Well, in turn, thanks for the effort shown in your post, and the work you did to include it! It's always a pleasure to work with users who are clearly motivated! – amWhy Mar 27 '13 at 19:49

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.

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so i wanted to understand generall principle.yes of course we should choose smaller intervals,but generally if at critical point function has smallest value then at this critical point's near interval,then this point is called local minimum,if maximum has,then local maximum,else saddle point,this is right yes? – dato datuashvili Mar 27 '13 at 19:38
@dato: that is correct. – Ross Millikan Mar 27 '13 at 19:50

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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