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So a bit confused when I see a question say: "Use the second derivative test to find all relative extrema"

When using the second derivative test are we not looking for concavity and points of inflection.

So far, in order to find relative extrema, the first derivative test would normally be used to find critical numbers and the critical numbers would then be evaluated on either side to determine in it was a relative maximum or minimum.

Could someone explain what is different when using the second derivative test?

Also, a bit confused on why the critical numbers are different in the first and second derivative.

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  • $\begingroup$ If $f''(x_0)>0$ then the function is convex in a neighbourhood of $x_0.$ Assume $f'(x_0)=0.$ Try to draw a function satisfying both conditions. Can $f$ have a maximum at $x_0?$ $\endgroup$ – mfl Nov 5 '16 at 23:15
  • $\begingroup$ No, it wouldn't it would have a relative minimum at that point would it not? since it's convex? $\endgroup$ – Siddart Fredrick Nov 5 '16 at 23:24
  • $\begingroup$ Yes. It has a relative minimum because it's convex. This is just the second derivative test for maximum/minimum. $\endgroup$ – mfl Nov 5 '16 at 23:26
  • $\begingroup$ Okay, but can't you find the relative max and min's but considering the increasing/decreasing intervals? Is using the second derivative test, just another way to do this? $\endgroup$ – Siddart Fredrick Nov 5 '16 at 23:31
  • $\begingroup$ No. It only works to say that a critical point is a maximum or a minimum. It doesn't say anything about increasing/decreasing intervals. $\endgroup$ – mfl Nov 5 '16 at 23:33
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The first derivative is the slope of the function, and the first derivative test is used to find the critical points, which are points where the derivative is equal to zero. The points are minimum, maximum, or turning points (points where the slope changes signs).

The second derivative is the concavity of a function, and the second derivative test is used to determine if the critical points (from the first derivative test) are a local maximum or local minimum. If the second derivative at a critical point is negative, then it is a local maximum, and if the second derivative at a critical point is positive then it is a local minimum. If the second derivative at the critical point is zero, then it says nothing about the concavity.

I hope this helps.

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I will illustrate the point considering the case for local maximum. To do so we first need a proper definition of local maxima.

Let $f$ be defined in a certain neighborhood $I$ of $c$. The point $c$ is said to be a point of local maximum of $f$ if $f(c) \geq f(x) $ for all $x\in I$.

As an example the function $f(x) =-|x|$ has a local maximum at $0$. Note that there is no talk about derivative of a function in the definition of local maximum. However using basic concepts from calculus one can find sufficient conditions to ensure that some point is a local maximum for a given function.

Thus one simple condition is that $f$ is increasing to the left of $c$ and then decreasing to the right of $c$. This will ensure that the values of $f$ on left as well as right of $c$ don't exceed $f(c)$ and $c$ is then a point of local maximum of $f$.

This is the basic idea (which is only sufficient but not necessary) which we use to check if a given point is a point of local maximum or not. Since this is based on concept of increasing / decreasing nature of functions, it is natural to assume that $f$ is differentiable in some neighborhood of $c$. And then by definition of derivative it can be proved that $f'(c) =0$ if $c$ is a point of maxima. Thus to find the point of maxima we only need to consider the points at which derivative vanishes. In what follows we assume that $c$ is a point with $f'(c) =0$.

And if derivative $f'$ is positive to the left of $c$ and negative to the right of $c$ then $f$ is increasing to the left of $c$ and decreasing to the right of $c$ and therefore $c$ is a point of maximum. This is what is usually called the first derivative test.

If we assume that $f$ is twice differentiable at $c$ and $f''(c) <0$ then this automatically ensures that $f'$ is positive to the left of $c$ and negative to the right of $c$ so that $c$ is point of maxima by first derivative test. In most cases computing $f''$ at a point $c$ is easier than verifying the change of sign of $f'$ around $c$. This is called the second derivative test. And you can see that it only ensures that the check mentioned in first derivative test is satisfied. But then it has a limitation that it can fail if $f''(c) =0$.

Based on actual function $f$ one should be able to figure which of the tests requires less effort and usually that particular test is chosen. You should observe that both the tests are sufficient and the first derivative test is more powerful of the two because it requires less conditions on $f$ and it can succeed in cases where second derivative test fails (when $f''(c) =0$).

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