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If I have the function $f(x)$, and the I take the first derivative, $f'(x)$, the point where $f'(x)=o$ is a minimum or maximum.

I learned in my Calc 1 class that to determine if this point is a min or max, I need to take the second derivative and see if it's $>0$ or $<0$ for all $x$.

Question: Why can't I just check a point $(f(x))$ on either side of $x$, where $f'(x)=0$? If the point I check is lower than $x$, than $x$ is a max, and if the point I check is higher than $x$, the point is a minimum. I can't figure out why I need to go through the work to find a second derivative.

Thoughts?

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    $\begingroup$ First derivative test is perfectly valid. $\endgroup$
    – user730361
    Commented May 18, 2021 at 14:41

3 Answers 3

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It is Ok, provided there is no max/min the the vicinity of the point of interest. this is why second derivative test is good.

Another, test is if $f'(a)=0$, then check the change of sign of $f'(x)$ when you go from left to right, if it changes from negative to positive, then $x=a$ is the point of min, otherwise max. Further, if it does not change sign, the there is no max or min.

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Some functions, like $f(x)=x^3$ have neither a max or a min when $f'(x)=0$, in this case, there is a saddle point at $x=0$ so just checking a single point nearby $x=0$ won't work. The second derivative test will catch this, because $f''(x)=0$, which means the critical point is not a max nor a min.

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  • $\begingroup$ Hi, interesting example. But wouldn't my proposal of checking f(x+a) and f(x-a) still work for f(x)=x^3 because x+a would result in a y value above f(o), and x-a would result in a y value below f(0). So without using the second d/dx, I was able to determine this is not a min/max point? $\endgroup$
    – findingmyway
    Commented May 18, 2021 at 15:33
  • $\begingroup$ As long as you check on both sides, and the function isn't too poorly behaved, then yes, your method will work, your professor might not be too happy with it though :) $\endgroup$
    – Jake Brown
    Commented May 18, 2021 at 15:39
  • $\begingroup$ Thank you! :) :) $\endgroup$
    – findingmyway
    Commented May 18, 2021 at 15:55
  • $\begingroup$ Not necessarily. The $f''(x)$ technique does not behave for e.g. $f(x) = x^4$ where there is an obvious min at $0$ but defiantly $f''(0) = 0$. $\endgroup$
    – Prime Mover
    Commented May 18, 2021 at 15:56
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The approach you suggest is exactly the way we were introduced to this subject back in the day when we were introduced to calculus at school. it's a valid approach, but it has to be pointed out that it does involve plenty of work.

On the other hand, if you are familiar enough with differentiation, taking the second derivative is often less work than specifically investigating what happens at $f(x-\epsilon)$ and $f(x + \epsilon)$, where (depending on the function under analysis) you may have difficulty in establishing the sign of $f$ at each of these points.

When I approach these problems, I routinely perform the second derivative test, because I find it easier than the $\pm \epsilon$ approach.

Unfortunately, the second derivative test does not always work. Take, for example, $f(x) = x^4$. The first derivative is $4 x^3$, which is $0$ exactly where $x = 0$, but the second derivative is $12 x^2$, which is also $0$ at $x = 0$.

So you can say the 2nd derivative test can be used except when $f''(x) = 0$ and in that case you have to use an approach based on investigating the sign of either $f(x)$ or $f'(x)$.

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