Now these are only local minima or maxima since we're just using derivative

information. And some students get things mixed up.

What do you do? Is it positive for a minimum or positive

for a maximum? Negative?

Which is it? I'll leave it to you to come up with a

way to remember how this goes. Now, why is it that the second derivative

test works? Well, as with the case of L'Hopital's

rule, there's a good reason. And Taylor series allow us to see that

reason transparently. Consider what happens at a critical

point, x equals a. If we Taylor expand f about that point.

Then what can we say? Well, because we have a critical point.

The first derivative vanishes. Then with that gone away and ignoring all

of the higher order terms that is terms of degree 3 and greater what do we see f

looks like quadratic, you have a constant term and them the leading term after that

is quadratic. In x minus a.

That means, ignoring the higher order terms, that the graph looks locally like

a parabola. The parabola would open up if the

coefficient in front of the second degree term, is positive.

That would open down, if the coefficient is negative.

Therefore we see that the sign of that second derivative at A, tells us whether

we have a local maximum or a local minimum.

Let's consider the simple example, sine squared to x times log of cosine of x.

This has a critical point at x equals zero.

What happens when we Taylor expand? Sine of x becomes x plus big O of x

cubed. We square that, and then multiply it by

the log of what? Well 1 minus x squared over 2 plus big O

of x to the 4th. We have a little bit more expansion to

do. Taking the sign squared term, and

expanding that out. And then, expanding log of 1 plus

something into, in this case. Negative x squared over two plus big 0 of

x to the fourth. Now when we multiply these three

expansions out, the leading order term is given by the product of x, x, and

negative x squared over 2. That is, we get a negative one-half x to

the 4th plus big O x to the 6th. And we can see why we didn't use the

second derivative test here, because it would've given us a 0, a fail.

However, we can conclude that this really is a local maximum at 0.

Because of this coefficient, negative one half in front of x to the 4th.

So what does that mean? Have we, have we just invented the fourth

derivative test? No, nothing that exciting.

We've just done Taylor Expansion. Instead of relying on specialized tasks,

rely on the principles. In this case, taylor expand to see what's

happening. Let's do a simple example.

Consider a square of side length l out of which cut the four corners we remove four

square of equal size and then fold at the internal edges to get an open topped box.

I want you to maximize the volume of that box with respect to what we remove.

If we remove larger squares we might get a smaller volume box.

If we remove smaller squares, then we might get a smaller volume Box.

What is the maximal volume? Well in this case, we better set x to be

a variable; in this case, the side length of the corners that we remove.

Then in this case, the volume is what? Well when we fold it up, the height of

the box is x. And the length of the base is quantity L

minus 2 X. So the volume is X times L minus 2 X

quantity squared. Multiplying that out gives 4 X cubed

minus 4 L X squared, plus L squared times X.

Now remember, x is the variable and l is a constant.

So that when we differentiate with respect to x, we get 12 x squared minus 8

l x plus l squared. This factors nicely as 6 x minus l times

2 x minus l. And we wanta factor that because we're

going to compute the critical points. Setting that directive equal to zero

solving get us two critical points. One that x equals L over 2 and the other

that x equals L over 6. We must classify these as maxima or

minima, or neither. At best we can, so we'll take the second

derivative, which we can compute to be 24x minus 8l.

Evaluating that second erivative at the ritical points gives us what...

At L over 2, it gives us 4 L. That's a positive number and that means

that we have a minimum or a local minimum at that point at x equals l over 6.

The second derivative is negative 4 L. That gives us local maximum.

Of course this fits with your intuition if X is equal to L over 2.

Then what is happening when you fold up that box?

Oh well you're getting something with volume 0.

That's clearly a minimum. We therefore conclude that the volume is

maximized. At x equals l over six but notice the

problem is asking us for the maximum volume not the value of x at which the

volume is maximized. So we need to substitute back in x equals

l over six into v. To get 2l over 3, that's l minus 2x,

squared, times x, or l over 6. Simplifying that gives us 2l cubed over

27. That is the maximum volume.