in section 4.1 our topic is maxima and minima so in many practical problems it's important to determine how large or how small a certain quantity can be so if we can translate a practical problem into some sort of mathematical description then we can often reduce that question down to finding the maximum or minimum of some function so in this section we're going to lay a foundation for the ideas that we will be using in solving optimization problems so in chapter four very broadly we're going to be looking at the derivative in applied settings and using the derivative as a tool so we've seen in chapter three the different ways that we can find derivatives of different types of functions and now we're going to begin to uh encounter applications of the derivative so chapter four very broadly now that we know how to find the derivative of a function we are going to use chapter four to go through various applications of the derivative so we need to introduce some terminology so here i have the graph of a function represented on some small window of its domain so you'll notice here that we have some labelings that we can apply to certain points on this graph so notice here that this point right here we are labeling as the global maximum and we're calling this the global maximum because of this entire segment of this curve this is the largest y value right somewhere close to six and on the other hand we have a global minimum here this global minimum is going to be the smallest y value on this entire segment somewhere at around negative three so we have the idea of a global maximum and a global minimum sometimes instead of global the word absolute is used so global and absolute are interchangeable in this context now we also have a concept of local maximum and local minimum so here this point is labeled a local maximum because it's the largest y value in a window around that point so if i were to draw a box around this region of this curve inside of that box this point is the largest y value and so we call that the local maximum on the other hand we have a local minimum so again think of drawing a box around this area within that box this point is the smallest y value and we would call that a local minimum so we have the contrasting ideas of a global max or a min or a local max or min so sometimes instead of local the word relative is used so the words local and relative are also interchangeable in this context so as an analogy think about your dorm and let's say that we separate the dorm by floors now if we were to measure the height of everyone on each floor certainly on a floor by floor basis there would be a tallest person and there would be a shortest person that would be the local maximum and the local minimum now think of considering the dorm as a whole so if we consider the dorm in its entirety certainly within the dorm there's a tallest person that would be the global maximum and there would be a shortest person and that would be the global minimum so the global maxima men considers the entire dorm the local max or min only considers an individual floor so mathematically we have the same type of situation a global max or min considers all y values whereas a local max or a min only considers the y values within a small window or within a small region of the curve so here's another picture demonstrating the concept of a local maximum or a local minimum so you see that individually within a small region around these points these points are either representing a maximum in that region or a minimum in that region so the maximum and minimum values of a function f are collectively referred to as the extreme values or the extrema of f so we might define these concepts of local maximum and local minimum so the number f of c is called a local maximum if it's the largest y value in a quote unquote window around x equals c now we have a way of defining a window mathematically but it's a little bit complicated so i leave it informal here likewise the number f c is called a local minimum if it's the smallest y value in a window around x equals c so notice that these definitions require this concept of local max or local min to be a number so that will have some significance in just a second when we look at a function that may not actually be defined at some point c but it has this behavior that we're seeing that mimics this concept of local max or local minimum so by this definition local extrema may not occur at an end point so i cannot have a local max or a local min at an end point again the concept requires us to be able to compare the y value to its neighbors so i have to be able to look at the curve within a window and compare points to the left of the window and points to the right of the window well that cannot happen on an endpoint so local max and local mins will not occur at endpoints and as i've mentioned some definitions use the word relative instead of local so relative and local are interchangeable so here's the graph of a piecewise function and what i'd like to do is label the extrema of this function so starting at this point you'll notice that this y value is the largest y value on the entire curve so there's no other y value on this the graph of this piecewise function that's larger than this y value and so that is going to be our global maximum so again we've got a global max here a local max cannot occur at an endpoint so we can only say that we have a global max at this point if we move here to this region around x equals negative 2 if i were to draw a box around the curve in this region and if i look within the box this point is going to be the smallest point or the smallest y value within that box and so this is going to be the local minimum so i'll abbreviate with just l min so within just this small area this is the smallest y value if we move to this next area if i were to draw a box around this region and if i look at the y values within the box so this y value is approximately 3 i would say if i move to the left the y values get smaller and if i move to the right the y values get smaller so what we have here is a local maximum so our definition for local maximum said nothing about continuity all it said is that we're comparing a number to its neighbors in a small window all right so we had no concept of continuity with these concepts of local max and local minimum so this being discontinuous here with a jump discontinuity is perfectly fine we have a local maximum if i look at this next area around x equals positive 1 again i want to compare the y values within that box so if i look to the right y values are larger if i look to the left y values are larger so what i have here is a local minimum and what you might notice in addition this y value is the smallest y value in consideration of the entire graph of this piecewise function so in addition we can label this as a global minimum so it's possible that a local minimum is also a global minimum and the same goes for maximum as well now at this next region if i draw a box around this this region of the curve you'll notice that we sort of have this behavior of a local maximum however there's no point here so this function the y value is undefined when x equals two so our definition for local max and local men required that we're talking about the number f of c and it being smaller or larger than the y values within a small window around the x value since there's nothing here though there is no y value to talk about being a local maximum so in this case we cannot label this as a local maximum so there's nothing that we can apply to this particular point here on this piecewise function right and lastly here at the end this value here is obviously zero this is an endpoint so obviously it's not going to be a local max or min now notice we've already exhausted the global maximum and the global minimum we cannot have more than one global max or global min so there's nothing that we can assign to this endpoint so our question going forward is how may we locate extrema so as you might have guessed the tool that we're going to use for locating extrema is going to be the derivative so i've given you the graph of a function here and at this point we're going to restrict ourselves to just talking about local maxes or local mins so i've got three points here x0 x1 and x2 at x0 we have a local maximum at x1 we have a local minimum what do we notice as occurring in terms of the derivative at these points you'll notice that the tangent line here in red is horizontal meaning the derivative at x0 and the derivative at x1 is going to be equal to zero so it seems that if we have a local max or a local min we should at least have in the first case we should have the derivative being zero now let's look at what happens at this point at x2 this is a corner we know the derivative does not exist at a corner so f prime of x2 does not exist however this is a local maximum so from this picture it seems that if we have a local max or a local min then we should have either one of two things the derivative is equal to zero at that point or the derivative does not exist at that point so this brings us to a very important theorem in calculus so the theorem states that suppose f attains a maximal or minimal value at x equals c then it must be that either f prime of c is zero or f prime of c fails to exist so this theorem gives us a first step towards locating max's and mins so it says that if we have a max or min then we must have either the derivative being zero at that point or the derivative failing to exist at that point however the converse is not true so we can't reverse this statement the fact that f prime of c equals zero or f prime of c fails to exist does not automatically imply a maximum or a minimum so when we find these points where the derivative is zero or non-existing we have candidates for maxes or mins we do not immediately have the conclusion of a max or a min all right so suppose f attains a maximal or minimal value at x equals c then it must be the case that either the derivative is zero or the derivative fails to exist at that point so a counter example is this function here y equals x minus a half quantity to the third power plus one the derivative y prime at this point x equals one half is going to be zero however if we look to the left values of y get smaller and as we look to the right values of y get larger so here the derivative is 0 at x equals one-half however this is not a max a and it's not a min so we cannot immediately conclude max or min from this statement or from this quality of the derivative being 0 or the derivative failing to exist at a point all right so these particular points have a name and we will name them this way so an interior point c contained within the domain of f is called a critical point if either f prime of c is equal to zero or f prime of c fails to exist so now going forward if we want to locate critical points we want to locate where the derivative is zero or where the derivative fails to exist and the critical points will give us candidates for potentially having a maximum or a minimum again we cannot immediately make that conclusion we'll have to develop some steps or some techniques later on in this chapter to go from critical point to the conclusion of max or min so again critical points give us candidates for extrema we cannot immediately go from critical point and make the conclusion of max or min so first we're going to focus on just finding the global extrema within a closed interval we'll have to come back later on in chapter 4 and talk about this idea of going from critical point to classifying that critical point as a local maximum or a local minimum so for now we're going to just focus on the global extrema within a closed interval that brings us to the extreme value theorem so the extreme value theorem says suppose f is continuous on the closed interval a b then there exists a minimum and a maximum y value on that interval so let's just think of some various ways that i could construct a continuous function on a closed interval so let's just take two endpoints and connect them continuously well certainly you see here that this would be a maximum and in this sense we're going to be talking about global maximum and here would be the global minimum so try for yourself take any two points connect them continuously in some way you're going to see that there's a maximum and a minimum value within that interval so how will we apply the intermediate value excuse me the extreme value theorem in this context if we have a continuous function on a b first we're going to calculate f of a and f of b next we're going to find the critical points of f on that interval say that perhaps we have critical points c1 and c2 we'll then calculate f of c1 and we'll then calculate f of c2 so the global max is going to be the largest of these values and of course the global minimum will be the smallest of these values so here we'd like to determine the global extrema of this function g of x equals 4x over x squared plus 1 on the interval 0 to 2. so first let's go ahead and calculate g of 0 you should notice that g of 0 is going to be 0 since we're going to have 0 over 1 and then g of 2 that's going to be 8 on the top and then 2 squared plus 1 5 on the bottom so now what i need to do is find the critical points that requires the derivative of course so the derivative by the quotient rule is going to be the derivative of the top four times x squared plus one minus the derivative of the bottom times the top all divided by the bottom x squared plus one quantity squared so there's our derivative by application of the quotient rule so for the critical points that's going to be where the derivative is zero and also where the derivative does not exist so first of all let's think about where the derivative is going to be zero if the derivative is going to be zero that's going to require the numerator to be equal to zero so we'll set the numerator equal to zero that's going to be four times x squared plus one two x times four x is going to be eight x squared so we'll set this equal to zero and solve so this is going to be 4x squared plus 4 minus 8x squared so this is going to be 4 minus 8x squared excuse me 4x squared minus 8x squared that's going to be negative 4x squared plus 4 equals 0. so what we'll have is solving for x we'll add the 4x squared over this is going to be 4x squared equals 4 and then dividing by one excuse me dividing by four will have two potential critical points x equals plus one and x equals minus one however we're only on the interval zero to two so i'm only interested in the critical point x equals one okay so we're only looking on the interval zero to two negative one clearly is not inside the interval 0 to 2. now for the derivative not existing if i look at the derivative the only issue for this derivative not existing is going to be the denominator equal to zero but you'll notice here that this denominator can never equal zero on the reals so if i were to solve this for zero that would if i were to solve this x squared plus one equal to 0 we would have x squared equals negative 1 which of course is not possible so we have no critical point coming from this case where the derivative is not existing so what i'd like to do now is just find the y value at this lone critical point of interest one obviously that's going to be four over two which is going to be two so what we see here is we see that zero is going to be the global minimum and eight over five is clearly less than two so two is going to be the global maximum so from the extreme value theorem we know that we have a max or a min on this interval indeed we have found the minimum is zero the maximum is two next is the function f of x equals the natural log of x squared plus two on the interval negative one to two again i want to find the global extrema first let's go ahead and calculate the y value at the endpoints f of negative 1 is going to be the natural log of 3 and then f of 2 is going to be the natural log of 6. so we want to find the derivative now the derivative of the natural log of x squared plus two is going to be one over x squared plus two times two x so we now need to find our critical points again the critical points are going to be where the derivative is zero and where the derivative does not exist so again setting the derivative equal to zero that's going to require the numerator to be equal to zero again if i have a quotient the only way i make a quotient zero is by the numerator being zero so we're going to have 2x equals zero of course our solution here is x equals zero now will the derivative fail to exist so the derivative will fail to exist if the denominator is equal to zero again you should notice that x squared plus two is not going to be able to be made zero on the reals so we have no critical point originating from the derivative not existing so we'll calculate this last y value f of zero and we have the natural log of two so which is largest and which is smallest it may help to think about the graph of the natural log function the graph of the natural log function looks like this it's an increasing function so since it's increasing natural log 2 is going to be our minimum and then the natural log of 6 is going to be our maximum