 Hello again and welcome to the sequel to the first screencast where we're looking at a family of functions, two-parameter family of functions f of x equals a times square root of b minus x squared where a and b are positive and we had done in the first screencast we found the domain of this function we found its critical values and we found where the function is increasing and decreasing and we found that every member of the family has a local maximum value at x equals zero of value a times square root of b. Now we want to move on to think about second derivative information namely we're going to find the intervals where this function is concave up and where it's concave down once we've done that we should be able to identify the inflection points very easily. This involves a fairly lengthy second derivative calculation so if you want to see that please hang on if you don't want to see that just scan through to the end where you get through the result. Now to set this second derivative calculation up I'm going to take the first derivative and just rewrite it slightly to make the derivative taking a little easier so f prime of x instead of writing it as a fraction with a root on the bottom I'm going to write it as negative a x times b minus x squared the quantity to the negative one-half power that's going to make derivative taking a little simpler. Now when I go to the second derivative of x again a and b are constants and particularly I have a multiplied constant right here in negative a so the first order of business for me is I'm going to pull out this negative a and take the derivative of what's left over so this is going to be the derivative of x times the quantity b minus x squared to the negative one-half so just to strategize before we go over to the next page and begin this is going to involve a product rule situation because I have two things multiplied together and the second thing is going to involve the chain rule because it's a composite function like I said this is a good recap of what you know about derivatives so watch onward if you want to see that. Okay so let's proceed with the derivative here the negative a is just hanging out here in front it's a multiply constant so the first thing I need to do is use the product rule to take the derivative of x times the expression in the one-half power so this is going to give me the derivative of x which is one times the second function which is b minus x squared to the minus one-half plus x times the derivative of b minus x squared to the minus one-half that's going to need the chain rule so let me switch to a different color to indicate the chain rule portion of this the chain rule I would take negative one-half times the outer function to the negative three-halves power with the original inner function put into it times the derivative of the inner function remember b is a constant so it differentiates to zero and the remaining derivative would be of negative x squared that gives me minus 2x and close that outer parenthesis let's do a quite a bit of cleaning up here to see what we can get done one thing I see is that the negative one-half times the negative two both of those completely divide off so and the one of course multiplying by one doesn't really do anything so I have negative a times let me write this all just the parenthesis portion here b minus x squared to the minus one-half plus let's see I have an x from here and then everything else there's another x down at the end so that's actually x squared and then I have this this expression here to the three-halves it's b minus x squared to the minus three-halves now let's try to simplify this expression as much as possible it may not seem like a good idea but it is actually going to be helpful to take the negative powers here the negative one-half and the negative three-halves and turn those into fractions with roots on the bottom so let's do that negative a times big parenthesis b minus x squared to the minus one-half is one over square root of b minus x squared plus I have an x squared over b minus x squared to the three-halves all right there we go now if I multiply the a the negative a through let's just do that real quickly I would have negative a over square root of b minus x squared minus or let me write it as this plus negative a x squared over again not really a root but negative or b minus x squared to the three-halves now what's going to be helpful for us here is to get common denominators on these fractions to kind of put them together so remember that I could just as easily instead of writing a square root here let me just erase that off I could have very easily rewritten this as b minus x squared to the one-half power now what's the common denominator the common denominator would be b minus x squared to the three-halves and I would get the common denominators by taking this first fraction and multiplying the top and bottom by something let me scoot that equal sign over if I multiply the top and bottom of this thing by b minus x squared to the first power then that's just multiplying by one that doesn't change the fraction but notice that's going to give me common denominators on the bottom because b minus x squared to the first times b minus x squared to the one-half is b minus x squared to the three-halves so on the next page I'm going to continue that calculation and we'll simplify this down completely so let's carry out the multiplication step that you see here on the top of the fraction on the left here I'm going to have negative a times the quantity b minus x squared and on the bottom as we discussed I'm going to have b minus x squared to the first times b minus x squared to the one-half that's b minus x squared to the three-halves now on the second fraction I have negative a x squared over b minus x squared to the three-halves okay so now we have common denominators I can add the numerators so multiplying this negative a uh through the first set of parentheses gives me negative a b plus a x squared this is all going to be over b minus x squared to the three-halves now look at what I'm adding I'm adding on a plus negative a x squared adding the numerators and so this is a really nice simplification these two guys just simply subtract off from each other and here's my final answer this is really worth the simplification step because it comes out to be nice and short negative a b over b minus x squared to the three-halves and that's your second derivative that's a fairly lengthy calculation obviously but every step is just basic derivative techniques carrying it out in algebra now what were we supposed to do with this second derivative we were supposed to use it to determine where the original family of functions was concave up and where it's concave down and then try to find inflection points now where is this function concave up well function is concave up whenever its second derivative is positive and concave down wherever its second derivative is negative so let's take a look at the second derivative expression here and just analyze its sign now on the bottom of this fraction in the denominator I have a quantity raised to the three-halves power a quantity raised to the three-halves power is always going to be positive because three-halves power means I'm cubing something and then taking a square root a square root never outputs a negative result and so that's always a positive number on the bottom on the top notice what happens here I've got a times b now at the very very beginning of this problem we said that a and b are both positive numbers so a times b is always positive therefore the numerator of this fraction which is minus a b is always negative so no matter what the x is no matter what the a is no matter what the b is and this function the second derivative of my family is always negative because I'm always taking a negative quantity divided by a positive quantity so what this tells me is that f is always concave down always concave down there's no value of x that makes f double prime positive f double prime is always negative so f is always concave down therefore there are no inflection points because an inflection point is where the concavity of f changes never changes so no inflection points so there's actually a fairly quick answer although we had to do a lot of algebra work to get to it so that pretty much fully specifies the family of functions here in terms of critical values increasing decreasing behavior concavity is very simple as a parting shot here let's go to geogibra and actually take a look at this family of functions and see if the behavior that we see actually fits with the calculations we made so as you can see here in geogibra I plotted the family of functions here using sliders to change the parameters here all the things that we calculated about this family turn out to be correct as you can see the domain of the function goes from negative screw to b to positive screw to b here's the value of screw to b right there and you can see it does actually fit as you change the b the domain gets larger or smaller the function is increasing from the left end of its domain up to zero and then decreasing until we get to the right end of the domain notice that the critical value there is at x equals zero and the local maximum value is always equal to squared of a times b you can change that either way you can see the parameters do different things to this function but some things always remain the same we also see that the function is always concave down sometimes increasing concave down and decreasing concave down but never concave up so there's no inflection points on the graph of this function and the great thing is we didn't need the visual to tell us this we found it completely through our use of the calculation tools that we know thanks for watching