The Second Derivative Test

Hey guys, this is 'Not Paul' scribing for Dec 2nd 2008. Today (or rather, yesterday as of the time this is being written), we continued talking about the Second Derivative Test and how we can use it. Oh and by the way the next scribe is Francis.

Okay, first, what is the second derivative test?

The second derivative test is basically a little trick that lets you know the concavity of your function.

If the value of the second derivative at a critical point (of the original function) is positive, then the original function is concave up at that point and is a local maximum. If it is negative, it is a local minimum.

To help me explain why this is, I have drawn a little squiggly:

Okay, so on this squiggly, we have atleast 4 critical points. For now, we'll deal only with one of the critical points to demonstrate the second derivative test. The easiest point to use is the local maximum at approxmiately (-1,1). I have labelled it C here.

Well, how is this related to the second derivative test? I'll tell, if you'd stop being so impatient.

The second derivative is the derivative of the first derivative right? And what is the first derivative, but the slope of the tangent line at a point in an interval? So our first function here, its on the interval [-3,3]. And the first derivative of our point c (c') is the slope of the tangent line at c.

As you can see, at our critical point c, c' is a horizontal line, which means the slope of c' (and the derivative of c) is 0. This is what the graph of our first derivative would look like (roughly):

Note the roots of the first derivative function at point c.

On a side note, as Benchmen pointed out we can find inflection points by looking for local maximum and minimums in the first derivative. When the first derivative function has a maxmimum or mimimum, that is where the slope of the original function is largest or smallest locally. That is basically the definition of an inflection point.

We can see that in the graph as the local maximums and minimums of the first derivative graph occur at inflection points on the original graphs. 

So anyway, our second derivative is just a derivative of our first derivative. Easy.

Yes, messy, I know, but I promise, this has a point.

See, if we take a look at my rough graph of the second derivative, we can tell three things:

a) At c, c" is negative

b) At c, c' is zero

c) The roots of the second derivative function are where we have inflection points

I'll make another point, g (for grue), and label it on the graph.

So, as we can see, where g" is positive, g' is zero and g is concave up.

So by this, we know a few things. First, we know that you can easily find the concavity of the function two ways:

By inputting the critical value (value being the x coordinate) into our second derivative function. If the output of the second derivative function is positive at that critical value, we know the original function is concave up. If it's negative, we know it's concave down.

Or by looking at the next value of the first derivative. Since we have value c which is a critical value, and we know c' is zero at c, if the value of the derivative function is negative at c + 1 or c + h where h is any small value (maybe its epsilon!) then the function is concave down (because the slope is becoming negative therefore the original function is decreasing). Conversely, if it's positive then it must be concave up.

Second, we know where the inflections points are (usually, or where they SHOULD be):

They are where you have roots on the second derivative, or where the second derivative changes from negative to positive

Orrr, where the first derivative has a local max/min (makes sense, as we know the derivative has a root at a critical point in the original function and second derivative is the derivative of the first derivative!)

So there you have it, thats what the second derivative test is about. Explained in a really long roundabout way.

Simple right?

So then, this stuff we did today should be a breeze:

We have a function f(x) = 4e^(-x^2).

We find the derivative of it with the chain rule (yes the chain rule!). Think of -x^2 as a function, so we have f(g(x)) = f(x) where g(x) = -x^2. So f(g(x)) = 4e^(g(x)). Simple! The derivative of 4e^(g(x)) is, well, 4e^(g(x)), and the derivative of g(x) is -2x. Multiply! Volia, -8xe^(-x^2) is our first derivative.

Since the only number than will make f'(x) = 0 is x = 0, then at x = 0 we have our critical value.

Then we derive again to get the second derivative.  Input the critical value into the second derivative equation, and f''(0) < x =" 0">

We apply the same process to the second questions. Remember! The derivative of ln(x) is 1/x.

Because of the +1 in the numerator, our function will never be undefined. So we only have to worry about the numerator. At x = 0, the first derivative is 0 so at x = o is where our critical value is.

To find the second derivative, we have to use the quotient rule. LO-DEE HIGH MINUS HI-DEE LO ALL OVER LO LO [In other words, the derivative of the denominator function times the numerator function minus the derivative of the numerator function times the denominator, all divided by the denominator squared].

Input x = 0 into the second derivative equation. It is greater than zero, so it must be a local minimum and the graph is concave up.

We also did some stuff that involved Grade 11 stuff, namely a function that looks like this:

[(x-a)(x-b)]/[(x-c)(x-d)]

Which serves as a lead in to our next topic of INFINITE LIMITS! I'd go into detail but Mr.K didnt post the slides for this, and it's 6:55am. 

Dont forget, homework was exercise 5.2 all odd numbered questions!

And lastly, the next scribe is not Joseph because he'll be away! Nah, the next scribe will not be Lawrence either, because that would be cruel.

Oh, didn't I say who the next scribe is already? I swear I wrote it at the beginning of the scribe post...

Here's something I whipped up as well for the whole scribe post badge thing.