 This lesson is on maximum and minimum values. You have already done lots of work on maximum and minimum values and even in this lesson you have done some work. But we will go on with definitions of new terms in this particular lesson. So the first term we're going to look at are critical points and critical values. The first thing we want to know is what are critical points? Any point in the form of a, f of a, where f prime of a is equal to zero or is not defined. And we know f prime of a is equal to zero when a graph looks like this and you have that point there or like this and you have that point or even like an x cube graph that looks something like that and that would be a critical value point where it's not defined could be points like on the absolute value function or on what I call the butterfly function some point where your derivative is not defined. What are critical values? These are the y values or f of a's where f prime of a is equal to zero are not defined. So critical points are the x and y values where critical values are just the y value. Let's have an example. We want to determine the critical points of the following function and you will see the function is a graph rather than an algebraic function. So we're going to look for where f prime is not defined or where f prime is zero. And as we look at this graph we can pick out this as a point where f prime is not defined. This point is where f prime is equal to zero. This point is where it's not defined. This point is where it's equal to zero and that point is where it is not defined. So the critical points in a list are zero, two, two, three, four, one, five, three and six, four where the function has the derivative that is defined and equal to zero are not defined at all. The next thing we have to learn about is the first derivative test. What is that first derivative test? The definition states if a f of a is a critical point and f prime changes sign at a then a f of a is either a local maximum or a local minimum. So the first derivative test involves finding a critical point and then testing whether it changes sign or not. Let's do an example of the first derivative test. Determine the local maximum of f of x is equal to e to the negative x squared. To determine this maximum we first have to find the critical point and to do that we have to take the first derivative. So we take f prime of x is equal to negative two x e to the negative x squared. Since the critical point is defined at f prime equal to zero or not defined we're actually going to take that zero option. And to solve for x we see that e to the negative x squared can never be equal to zero but we know that x can be equal to zero. So our critical point is when x is equal to zero. Now to use the first derivative test we write down f prime test and we want to know if the derivative changes sign at x is equal to zero. So the quickest way to do this is with a number line type of test. Put a zero in and if we put a number that's greater than zero like one into our f prime we see this becomes negative and of course this will always be positive therefore this is negative. If we put a negative number in like negative one this part of the prime will be positive and of course again this will be positive so it will be positive. So we know our function is going from increasing to decreasing which means we have a local max at x equal zero. So in this example we've employed again our critical point and then showed how the f prime test is used. Let's go to another definition that we need to know what is the second derivative test. Well the first part says if f prime is equal to zero and f double prime is greater than zero then f has a local minimum at a. Okay that means we have something that looks like this where f prime is equal to zero because as that horizontal tangent line f double prime is greater than zero which means this is concave up then we have that local min. Second part of this states if f prime is equal to zero and f double prime is less than zero then f has a local maximum at a. That means again we have the horizontal tangent line right there and then it's concave down that means we have a local max. Now we have a third item to look at if f prime is equal to zero and f double prime is equal to zero we really can't tell anything at this point because we don't know whether it's a point of inflection or whether it is something else. So if both the prime and the double prime are equal to zero we cannot tell what's going on. Let's do an example on the second derivative test. Let's continue with our f of x is equal to e to the negative x squared and take a second derivative on that. So our first derivative f prime of x was equal to negative two x e to the negative x squared. If we take the second derivative f double prime of x we get negative two. We have to use product rules so we'll have x times negative two x e to the negative x squared plus e to the negative x squared. Cleaning this up we have negative two e to the negative x squared times the quantity negative two x squared plus one. To use the second derivative we will have f double prime at zero where we determined our critical point. So f double prime at zero is equal to negative two for the negative two and then e to the negative x squared is one times the quantity zero plus one which equals negative two. Because f double prime at zero is a negative number we have a function which is concave down at that point which means we have a local maximum. And this concludes our lesson on maximums and minimum points.