okay welcome to section 4.1 maximum and minimum values so in this section we're gonna examine how derivatives affect the shape of the graph of the function and how they help us locate maximum and minimum values so some definitions first absolute maximum we say a function f has an absolute maximum at C if F of C is greater than or equal to f of X for all X's that are in the domain of F meaning there are no function values or no y values that are larger than F of C on the functions domain absolute minimum a function has an absolute minimum at K if f of K is less than or equal to f of X for all X's that are in the domain or an element of the domain of F local or relative maximum a function f has a local maximum at be if SS B is greater than or equal to f of X when X is near B what exactly does Emir mean another version of this definition you might see would say in an open interval containing B so don't specify exactly what the distance is but there needs to be some open interval that contains B where your local max occurs and local or relative minimum a function f has a local other should say minimum at M if F of M is less than or equal to f of X when X is near M or we could say in an open interval containing I'm gonna talk more about that in a minute okay so let's look at a graph here and see if we can apply the definitions and locate all of the absolute and local maxes and mins or extrema sometimes you'll see them referred to as extrema if it's a max or a min okay so first things first what pops out at you well right here that's SB that's a local max right it's the largest value of the grass near 4 on an open interval containing B it's not the largest function value I see on the entire graph correct so it's not gonna be an absolute max okay let's come over here to the next point F of C is a local min right it's the smallest function value on this little neighborhood or open interval um F of D that's nothing notice you might think oh it looks like a man or maybe you think it looks like a max that's the problem so it's the smallest Y value for function values to its right but it's the largest function values if you're approaching from the left so that's nothing then so sorry okay here at E f of E is a local max again it's the largest function value in an open interval containing e moving on F of F also looks like it's a local min but that's not all isn't it the smallest function value on the entire graph that also makes it an absolute min and then moving on here F of G well it can't be a local max because there is no open interval containing G it's an endpoint but it is the absolute max that's the function and this leads us to a very important point because that's why I also noticed us as a was nothing right it's not a local min or something like that because an endpoint cannot be a local min or max and if you think of that definition that a local min or max has to have an open interval around it or containing it then it basically has a neighborhood of function values around it so it have to have neighbors to the left and neighbors to the right meaning it cannot be an endpoint so let's write that down note endpoints are not considered for local Max or min values all right and points can be absolute Max or min values but never local just think they have no neighbors on one side okay let's move on to another example here sketch the graph of f to find the absolute and local maximum values of f another way you might see this problem worded is to find the local extrema that means the same thing that's just a way of saying Max and min at the same time okay so let's look at this function here we have a parabola right and it's been shifted one unit to the left and one unit upwards and then our domains restricted from negative two to two so here's my x-axis here's my y-axis I'll scale the x-axis by two since we don't have to sketch too much there but the Y values are going to increase a little more rapidly so one two three four five six seven eight nine ten perfect all right so F of negative two let's see what we're gonna get so F of negative two that's gonna be one plus negative two plus one squared so that's gonna be two so F of negative two is two F of negative one that's going to be one plus zero squared so that's one and then F of zero that's going to be one plus one squared so that's two F of one that's gonna be one plus two squared so five and then F of two is gonna be one plus three squared which is ten but notice here X is strictly less than two so what does that mean well I'm gonna plot an open circle at ten okay so here's my parabola no arrows right since the domains restricted beautiful all right so let's use this graph now we're gonna find the absolute and local extrema so I'm gonna list out all four options so absolute Max absolutely min any local maxes and you can always have more than one of absolute or local so they don't have to be unique all right do we have an absolute max do we have a largest function value no in this case there is not one since we have an open circle here at ten there is no function value that we could say is the largest and you might say well what about nine point nine well I can always find something larger that's on the graph how about nine point nine nine oh well why doesn't that work for the absolute max but what about nine point nine nine nine so this could go on forever so there's no absolute max because we have an open circle at ten if they're an absolute min is there a value on the graph that's the smallest function value on the entire interval well in this case yes right here so you want to write it out as follows you'll always want to identify where it occur so you will write F of negative one indicating that it occurs when X is negative and that equals one so that men value is one are there any local maxes I don't see any right so there's none are there any local mins well our absolute men also happens to be a local min right isn't it the smallest Y value on an open interval of values near negative one so we would say F of negative one which is one is also a local min okay so there's some exercises in the homework where you're gonna have to graph a function and then identify the following and not in every case will it have absolute and local extrema or it might have more than one for some cases all right moving on next idea is the extreme value theorem if f is continuous on a closed interval from A to B then F attains absolutely an absolute maximum value F of C and an absolute minimum value F of D where C and D are elements of the interval from A to B so notice we have two conditions in order to apply the extreme value theorem we require continuity of our function as well as a closed interval and then what the extreme value theorem tells us is that I'm guaranteed to find an absolute Max and an absolute min on this interval well why is this so powerful let's think of another parabola perhaps okay just our basic parent function y equals x squared okay if I look at y equals x squared well it's continuous right it's a polynomial but I'm not going to consider a closed interval from A to B I'm just gonna consider y equals x squared on its entire domain of all real numbers do I have an absolute max no right there isn't one because the function extends upward forever so infinity cannot be an absolute max we have to have a finite function if they're an absolute min yes right here at the vertex so f of zero which is zero so I haven't applied the extreme value theorem because I didn't consider the function on a closed interval so what if I do that now notice what changed this so y equals x squared and I'm only gonna consider it on the interval from A to B it doesn't matter where a and B is where they are okay so here's my parabola and say I'm gonna put a here B here all right now I'm only gonna consider the function on the interval from A to B do I have an absolute max only looking at the portion of the parabola from A to B why yes right here at the endpoint so f of B or B squared is my absolute max do I have an absolute min yes still at F of zero so basically what the extreme value theorem tells us is if we have a continuous function and we restrict the domain to a closed interval we can find both an absolute Max and min we'll see why this is so powerful in just a moment now for Moss theorem if F has a local Max or min value at C and if F prime of C exists then F prime of C is equal to zero and you may have noticed this before basically if you have a local min or a local Max and you try to draw a tangent line at that min or max value what is the slope of that tangent line well it's always horizontal right so the slope is going to be zero and that's what from our theorem tells us great so now this leads us to a very important idea of critical numbers and as long as C is in the domain of our function don't forget that part we say C is a critical number if one of the following two conditions is true F prime of C equals zero or if F prime of C does not exist and nope if F has a local Max or min at C then C is a critical number of F so basically our local extrema are a subset of the critical numbers if we have a local Max Ehrmann then it occurs at a critical number of the function the reverse statement is not true not all critical numbers or local maxes or mins but we'll talk about how to figure out which ones are later on in the chapter okay so let's look at an example now find the critical numbers of H of P which equals P minus 1 over P squared plus 4 so remember our criteria for critical numbers are they occur where the derivative equals 0 or where the derivative does not exist so what this means is we need to find H prime of P and then solve for where H prime of P equals 0 as well as where H prime of P does not exist ok so let's go ahead and find the derivative first H of P is P minus 1 over P squared plus 4 so we're going to need to apply the quotient rule here we go so H prime of P is going to equal we have low D high so low just the denominator D hi Devon is 1 is 1 minus high P minus 1 d low derivative of P squared plus 4 is 2 P over lo low so P squared plus four squared that needs to be parentheses okay now let's clean this up so we're gonna have P squared plus four minus now this is gonna distribute 2p squared plus 2p all over P squared plus 4 quantity squared beautiful and then combining like terms I have negative P squared plus 2p plus 4 over the quantity P squared plus 4 squared so there's our derivative now in order to find the critical numbers I need to do two things first find where H prime of P equals zero as well as where H prime of P does not exist I recommend you write these two conditions out first so you don't forget okay well the derivative is gonna equal zero when the numerator of H prime of P equals zero right so we're just gonna set the numerator equal to zero so negative P squared plus 2 P plus four is zero so we need to apply the quadratic formula let's just hit this thing with a negative one though so you never want that leading coefficient to be negative P squared minus 2p minus four is zero so that means P that doesn't factor does it no it's a quadratic formula time it is T equals opposite of B plus or minus square root B squared minus 4 ay C all over 2 so this is gonna be 2 plus or minus the square root of let's see that's gonna be 4 plus 16 over 2 so that's 2 plus or minus rabb 20 over 2 which is 2 plus or minus 2 rad 5 over 2 which is 1 plus or minus Rab 5 ok so those are two critical values right there 1 plus Rob 5 and 1 minus round 5 now to find where the derivative does not exist that would happen if the denominator was equal to 0 so I'm going to set the denominator equal to 0 so we set P squared plus 4 squared equal to 0 well taking the square root of both sides that would mean P squared plus 4 equals 0 which means P squared would equal negative 4 but that's not possible so this case doesn't yield any critical values but I do have some from where H prime of P equals 0 so those are the only two critical numbers that we found 1 plus or minus R outside all right nice job that concludes part 1