Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College Concavity. The graph of a differentiable function Y equal F of X is concave up on an interval I if F prime is increasing on I. The function is concave down on an interval I. If the first derivative is decreasing on I, the second derivative tests for concavity. Let Y equal F of X be twice differentiable on an interval I. If the 2nd derivative is greater than 0 on the interval, the graph of F, the original function over the interval I is concave up. If the 2nd derivative is less than 0 on I, the graph of F / I is concave down. A point where the graph of the function has a tangent line and where the concavity changes is a point of inflection. So if we look at this, we have concave up here and then at this point we change to concave down and there's a tangent line there. At a point of inflection, CF of C. Either the 2nd derivative IS0 or the 2nd derivative could fail to exist If we looked at a function that went like this, that's concave down changing to concave up at this location, and that would not have a function because it would be a vertical asymptote there. So that would be a second derivative that fails to exist. Should the first derivative say it fails to exist and hence the 2nd derivative can't exist either? If F prime of C = 0 and the 2nd derivative is less than 0, then F has a local Max at X = C So if the first derivative equals zero, IE we have a Max or a min, and if the 2nd derivative is less than 0, if it's less than 0, we know it's concave down South. F has to have a local Max at that location, the first derivative zero and the 2nd derivative is less than 0. If the first derivative equals zero and the 2nd derivative is positive, IE we have a 1st derivative 0 and a second derivative positive, then we have a local min because that first derivative being positive means it's concave up at X = C If F prime of C = 0 and the 2nd derivative also equals 0, then the test fails. It may have a Max, it may have a min, or it may be neither. It just doesn't give us any information if both the 1st and the 2nd derivative are equal to 0. When we're graphing Y equal F of X, we're going to identify the domain, so figure out which X values actually are defined. We're going to identify any symmetries looking for even and odd. Remember F and negative X. If it equals F of X, it's even. If it equals the opposite of F of X, it's odd. We're going to find both the 1st and the 2nd derivative. The first derivative is going to tell us those critical points. Max's and Min's second derivative is going to tell us points of inflection. We're going to find where the curve is increasing and decreasing off of the first derivative. We're going to find where the curve is concave up and down for the 2nd derivative. Identify any asymptotes, find your X&Y intercepts, and then sketch behavior of some graphs. The differentiable smooth connected graph may rise and fall. If we have our first derivative greater than 0, it rises from left to right, but it might be WAVY. If the first derivative is less than 0, it falls from left to right, but it also may be mate WAVY. Second derivative greater than 0, it's concave up throughout, no waves. Graph may rise or fall. Second derivative less than 0 throughout, concave down, no waves. The graph might may rise or fall if the 2nd derivative changes signs at an inflection point. We go from concave up to concave down or concave down to concave up. Y prime changes sign implies the graph has a local Max or local min. So if the Y prime changes from positive to negative or negative to positive, we have a local Max or local min. Y prime equals zero and the 2nd derivative is less than 0. There's a point that has a local Max. If the first derivative is 0 and the 2nd derivative is greater than 0, we have a local min. Looking at some examples, we start with X to the 4th over 4 minus two X ^2 + 4. So we know our domain here is going to be all reals. If our domain is all reals, then we need to find our first derivative. So we're going to get X ^3 -, 4 X. We also need our second derivative three X ^2 -, 4. Now if we can, it will help us to factor. So we get X * X + 2 * X -, 2. And here we could think of this as difference of two squares, root three X - 2 and root three X + 2. So if we thought about putting these on a number line, we would have -2 zero and two. We'd also have our -2 root 3 / 3 and our +2 root 3 / 3. If we look at this first derivative, let's choose a number to the right of two, say 5-10. A thousand doesn't matter. Is it a positive number or a negative number and it's going to be a positive? The multiplicity at two right here is odd. So if it's positive on one side, it's going to be negative on the other. At zero, our multiplicity is odd. So if it's negative on one side, the Y values on the other would have to be positive. And at the -2 right here, it's also odd. So if it's positive on one side, it's going to be the opposite or negative on the other. So when we look at this next piece, the 2nd derivative, if I choose something way out to the right, a hundred, 1000 a million, we know that to the right of two root 3 / 3, it's going to be positive AT2 root 3 / 3 because of the degree, the degree being one, it's going to be negative. And then from -2 the oops back up a little bit at -2 root 3 / 3 because the multiplicity is one, it's got to change back. So this first derivative tells me were decreasing, but we're also concave up from negative Infinity to -2 from -2 to -2 root 3. We're increasing because the first derivative is positive, and we're concave up because the 2nd derivative is positive. From -2 root 3 / 3 to 0. We're still increasing because of that first derivative still being positive, but now we're concave down from zero to two root 3. Those are both negative, so we're decreasing based off the first derivative and we're concave down from 2 root three to three. That first derivative is negative, so we're still decreasing, but now we have positive for the 2nd derivative, so we're concave up. And finally, from 2:00 to Infinity, we're increasing from the first derivative and we're concave up from the 2nd derivative. So if we thought about looking at this graph, actually let's find our X intercepts and Y intercepts. If we stick zero in 4X, we get Y being 4. The X intercepts maybe aren't so easy to find, so let's actually start with just the Y intercept. So here we know 04 and we know that -2 off to negative Infinity. We're going to be decreasing, but we're going to be concave up. So we're going to be decreasing. I'm going to actually make this a rough sketch. So we're going to be decreasing concave up until we hit -2 at -2. We're now going to be increasing but still concave up. So that makes this -2 be a local min at -2 root 3 / 3, we're increasing concave up, but now we're going to continue increasing, but now we're going to be concave down. So that's going to make this a point of inflection at zero. We're going to be starting to decrease, but we're still concave down. So if we're decreasing now, instead of increasing but concave down. This is a local Max AT2 root 3 / 3. We changed from concave down to concave up, but we're still decreasing. So this is a point of inflection at 2:00, we're going to change from decreasing to increasing, but we're still going to be concave up. So this is going to be a local min. So this is a rough graph of what this curve would look like. Looking at another example, if we distribute this out, we get 9 fourteenths X to the 7 thirds -9 halves X to the 1/3. So we need to find the first derivative which is going to be 3 halves X to the Four Thirds -3 halves X to the negative 2/3. Now we want to combine that and get that as a common fraction. So if I have X to the Four Thirds -1 / X to the 2/3, I just factored out the common three halves. We would have to multiply that first term by X to the 2/3 top and bottom. So we'd have X ^2 -, 1 / X to the 2/3, and X ^2 is going to be X + 1 * X -, 1 all over X to the 2/3. We want to also find our second derivative, which is going to be two X to the 1/3 plus X to the -5 thirds. Getting that to be over a common fraction, we're going to end up with two X ^2 + 1 / X to the five thirds. So looking at the first derivative, we're going to have it be 0 and or undefined at -1 zero and one. If I think way out to the right of 1567 a hundred and stick it into this equation, we're going to get a positive number. So the first derivative to the right of one is going to be increasing the first derivative between zero and one because the multiplicity on that one is odd. We know it's got to be negative. Now the multiplicity on the zero, that 2/3. So whatever goes in is going to get squared. It's going to be cube rooted also. But the square is the important part. So if it's negative on one side, it's got to be negative on the other. If you're having a hard time with that, think about putting in maybe a number like negative 1/2. If I put in negative 1/2, negative 1/2 + 1 is positive, negative 1/2 -, 1 is negative, but negative 1/2 ^2 is positive. And then when we cube root of positive, it's still positive. So in between -1 and one, we know the original function will be decreasing at -1. The multiplicity there is one. So we're going to change from a negative on one side to a positive on the other. Looking at the 2nd derivative, it's never undefined. No, sorry, it's never zero. It's undefined, however, at 0. So when we look to the right with the 2nd derivative, say 123, we're going to have positive values. If we look to the left, however, that's a 5, 5 thirds. If I take a negative value to the fifth power, it's going to be negative. Then if I cube root a negative, it's negative. So the top is always going to be positive. The numerator is always +2 * X quantity squared plus one. So two X ^2 + 1 is always a positive number, but that X to the 5 thirds is going to be a negative for all negative values. So that's going to come in as a negative. So now when we look at this, we know that in this first interval, the original function is going to be increasing, but we're going to be concave down. And -1 to 0, because the first derivative is negative, we're decreasing. The 2nd derivative is also negative, so we're concave down. From zero to one. The first derivative is still negative, so we're decreasing, but now the second derivative is positive, so we're concave up. And from one to Infinity, we're going to be increasing because the first derivative is positive and we're going to be concave up. So our graph, a rough sketch, we're going to be increasing, but concave down at -1 we're going to start decreasing and still stay concave down. So we know that that means that this has got to be a local Max at zero. We're going to keep decreasing, but now we're going to change to concave up. So for changing to concave up, that's got to be a point of inflection. So we were decreasing, but we changed to concave up. OK, not the scale here by any means. Decreasing, decreasing, decreasing, concave up. At this next point, we're going to be increasing but stay concave up. So that's going to make it a local min down there. Now, if we actually wanted the points, we would stick in these X values -1 zero and one back into the original equation. So -1 if I plug it in and chug it through, I get 27 sevenths as my local Max. My point of inflection is easy. Zero goes in, zero comes out, and for my location at one I believe I'm going to get -27 sevenths. If I put in one here, I get 9 fourteenths -9 halves, 9 fourteenths -9 halves, 963 halves, oh, 63 fourteenths. So that's going to be -54 fourteenths, so -27 sevenths. Let's look at an example that has some trig in it. So if we have Y equal tangent X minus 4X and we're giving on the interval negative π halves less than X less than π halves. So we know that we're going to have end points negative π halves and Pi halves. Actually we're going to have asymptotes there on the original graph. So our first derivative is going to equal secant squared X -, 4. Our second derivative, we're going to have to use the chain rule. So we bring down the two and take the secant X to one less power. But then we also have to take the derivative of the inside or the secant X. So our second derivative is going to be two secant squared X tangent X when we set the first derivative equal to 0 and solve. We're going to get secon X -, 2, secon X + 2. So secon X equal to X equals secant inverse of two, and X equals secant inverse of -2. And because we know our unit circle, we know that that's going to occur at π thirds and negative π thirds. Our second derivative, we're going to set equal to 0 and we're going to get 2 secant squared X tangent X. So tangent X is 0 when X = 0. And then we would also need to think about where is secant 0 because or not secant, yes, cosine. If we thought about putting this into an improper fraction, we'd get 2 sine X over cosine cubed X. So that derivative would be undefined when cosine X is zero. Well, that occurs at X equal π halves and negative π halves. And that's outside of our restriction. So we don't have to worry about that. We actually we don't have to worry about that because that's where it's undefined. So when we look at this, we're going to think about our first derivative and then our second derivative on a number line. They have negative π thirds here and π thirds here. And we know that we're going to just be talking about in between π halves and negative π halves. So if we put zero in to the first derivative, secant of 0 is 11 squared is one, 1 -, 4 is -3. So in between negative π thirds and π thirds, we're going to have a negative π thirds. Because the term was an odd multiplicity, we're going to change from negative to positive. At negative π thirds it was also an odd multiplicity, so it's going to change there. Our second derivative occurred at 0, so to the right of 0 in between 0 and π halves, let's say π force, maybe two secant squared π force times tangent π force. We know that's going to be positive because we actually know all the trig functions in the first quadrant are positive. And if we look at negative π halves to 0, the secant's positive there, but the tangents negative. So when we multiply, we'd get negatives there. So we know that the first derivative being positive means it's increasing, but the second derivative being negative means it's concave down. In between negative π thirds and zero, we're both negative, so we're decreasing and we're still concave down. In between 0 and π thirds we're decreasing, but now we're going to be concave up because of the positive. On the 2nd derivative π thirds to Pi halves, we're going to be increasing and we're going to be concave up. So doing a rough sketch of this one, we're going to have increasing but concave down. At negative π thirds, we're going to be decreasing and concave down. So that makes this location a local Max at zero. We're going to still be decreasing, but now we're changing to concave up. So we're still decreasing, but now we're concave up. So this is a point of inflection at 0 at π thirds we're going to be decreasing and up concave up. Did I miss an interval? 1234? So we went decreasing concave down, and then we went decreasing in concave up, and that π thirds we're going to be increasing and still concave up. So at π thirds we're going to have a local min. So that π thirds is a local min. The zero is a point of inflection and the Pi thirds negative π thirds is a local Max. To actually find those points, we would just stick into the original F and negative π thirds, F of 0 and F of π thirds. We can't include those endpoints because on the original it wasn't equal at those locations. So if we put in negative π thirds, we get negative root 3 + 4 Pi thirds. If we put in zero, we get out zero, and if we put in π thirds we get root 3 -, 4 Pi thirds. Looking at a few more examples, another trig 1. If we have Y equal to cosine X minus root 2X on an interval negative π less than or equal to X less than or equal to three Pi halves, our first derivative is -2 sine X minus root 2. Setting it equal to 0 and solving, we're going to get sine X equaling -1. No, we're not setting it equal to 0 and solving. We're going to get root 2 equal -2 sine X. So sine X equal negative root 2 / 2. We have to pay attention to our given interval. So sine X being negative, root 2 / 2 is going to occur when X is -3 Pi fours. Negative Π fours end up five Pi fours. If we think about our second derivative that's -2 cosine X letting it equal zero, we know that cosine X is 0 at negative π halves, Pi halves, and three Pi halves. So this one's going to be more complex when we look at this, the first derivative put -3 Pi force out here, negative π force somewhere in here, 5 Pi force somewhere down here. So if we're to the right of five Pi force to three Pi halves, we have end points here that's important with the trig ones. So in between 5 Pi force and three Pi halves, if we choose a number in between there, we're going to stick it in to the first derivative equation and we're going to figure out whether it's positive or negative. So if we stuck in four Pi thirds, sine of four Pi thirds would be negative. Negative times a negative is a positive. And then if we take away root 2, we're going to end up being a positive value there in between negative π force and PI5 Pi force. I like zero personally. Sine of 00 negative 2 * 0 minus root 2 is a negative value. Negative Π force to -3 Pi force. If we choose negative π halves, maybe we get a -1 a -2 * -1 is +2 minus root 2 is still going to be a positive. And from -3 Pi force to negative π if we use -5 Pi 6, we would get -3 Pi force. We're going to get negatives. So if we thought about our number line, this is five Pi force, six Pi force. Nope. Yeah. OK, so there's our positive, negative, positive negatives for the first derivative. Now we have to do the same thing for the 2nd derivative. The 2nd derivative is going to have negative π halves. Just going to occur in here, and then we're going to have Pi halves, which is going to occur somewhere in here. And we're going to have three Pi halves, which is our endpoint over here. So Pi halves to three Pi halves. Let's choose π maybe if I put π into here, cosine of π is -1 negative 2 * -1 would give us positive in between negative π halves and Pi halves. Let's choose zero -2 times cosine of 0 which is one would give us a negative and then from negative π halves to negative Pi -3 Pi force. Maybe the cosine of -3 Pi force is a negative and a negative number times a negative number is going to make it a positive. So first derivative is negative, so we know it's going to be decreasing. Second derivative is positive so it's going to be concave up. This next interval we have both of them positive, so it's increasing and up the next one, the first derivative is positive, so increasing. The 2nd derivative is negative, so it's down. The next interval they're both negative, so it's decreasing and it's down. The next interval, the first derivative is negative, so it's decreasing, but the second derivative is positive, so it's concave up. And the last interval it's going to be increasing and up because they're both positive. So from negative π to -3 Pi force, we're going to be decreasing, but we're going to be concave up. At -3 Pi force we're now going to be increasing and still be concave up. Negative Π halves, we're going to be increasing still, but now we're going to be concave down. So we're going to be increasing still, but concave down. At negative π force, we're going to go decreasing and still be concave. Down Π halves, we're going to still decrease, but now we're going to go concave up. That's a smooth continuous curve, concave down to up. Five Pi force, we're going to increase and we're going to go up. So at -3 Pi force because we were decreasing but concave up and now we're increasing. This is going to be a local min. So we're decreasing concave up. And at this point we're going to increase and be concave up. At negative π halves we're still increasing, but we're changing our concavity from concave up to concave down. So this is a point of inflection negative pie force we're going from increasing to decreasing, but we're both, they're concave down on both. So that's a local Max at pie halves, we're going decreasing down to decreasing up. So this is another point of inflection, and decreasing up to increasing up is going to make it a local min. Now on this problem, unlike the last one, we actually are including the endpoints. So this one is also going to be a local Max at three Pi halves. And down here at negative π we're going to have a local Max Rough sketch. So if you were asked on what interval is it increasing, you could tell me by just looking at the number line we did up here. So it's increasing from -3 Pi force to negative π force, five Pi force to three Pi halves. It's decreasing negative π to -3 Pi force, negative π force to five Pi force. It's concave up, negative π to negative π halves, Pi halves to three Pi halves, concave down negative π halves to Pi halves. What if we are given the SEC, the first derivative, and asked to figure out what this graph would look like? If we're given the first derivative, we're going to then going to need to find the 2nd derivative, so the second derivative here. If we multiplied it out, we'd get four X -, 3 X squared, pulling out the X that they have in common. So we know that the first derivative is going to have zero and two. So if we think about looking far off to the right, say 456, we get a positive squared, and 2 -, 5 would give us a negative. So to the right of two, we're going to have a negative for the first derivative at 2:00 because its multiplicity was odd. We know it's going to go from negative to positive. And at 0, because it's multiplicity is even, it's going to stay the same sign on both sides. So if it's positive on one side, it's positive on the other. At the 2nd derivative, we have zero and we have three, we have 4 thirds. So we're going to put 4 thirds in here. So the second derivative, if we're to the right of Four Thirds, say two, say 10, whatever you'd like, we'd get a positive number times a negative. So these are all going to be negative out this direction. In between -4 thirds and zero, we would have put in the number one, maybe positive times a positive. So a positive, we can also base that off the multiplicity right here it was odd. So if it's negative on one side, it's positive on the other, the 0 odd multiplicity again. So we're going to have negatives. So our original function is going to be increasing and concave down, then it's going to be increasing and concave up, increasing and concave down, and finally decreasing and down. Literally reading this off of the 1st and 2nd derivative. If it's positive for the first derivative, it's increasing. If it's negative on the 1st derivative, it's decreasing. If it's second derivative is negative, it's concave down. If the 2nd derivative is positive, it's concave up. So we're going to have increasing concave down until we hit zero. Then we're going to be increasing in concave up. So this is a point of inflection. At 4 thirds, we're still increasing, but we're changing from concave up to concave down. So this is going to be another point of inflection. And then at 2:00, instead of increasing, we're now going to start decreasing and still be concave down. So this is going to be a local Max. So we have two points of inflection and a local Max. For this example, let's do one more. If we're given the 2nd deriv or the 1st derivative, we have to find the 2nd derivative and the second derivative. Here we're going to need to do the product rule. So the derivative of the first two X -, 2 times the 2nd plus the derivative of the second bring down the two that X -, 5 to the one less power, and then the derivative of the inside, which in this case is one all times the 1st. I'm going to factor anything I can factor. If we look at this, we could take a two out of the 1st, so we could take a two and an X -, 5 out of both the terms. If we take a two and One X - 5, we get left an X - 1 * X - 5 in the first term and an X ^2 - 2 X in the second term. So this 2nd derivative is going to be 2 * X - 5. Foiling stuff out, we'd get X ^2 - 6 X plus 5 + X ^2 - 2 X, so 2 * X - 5 two X ^2 - 8 X plus 5 S For the first derivative, we're going to have 02 and five. If we think really far off to the right past five, we're going to get positives. The multiplicity at 5 is even, so if it's positive on one side, it's positive on the other. The multiplicity at two is odd, so it's going to turn to negative. And the multiplicity at 0 is also odd, so it's going to turn into positive. Now for the 2nd derivative. We get 5 from here, but we actually have to figure out this one and it doesn't factor. So we're going to use quadratic formula. X equals the opposite of B plus or minus the square root b ^2 -, 4 * a * C all over to a. So if we simplify that up, we're going to get 4 plus root 6 all over two and we're going to get 4 minus root 6 / 2. So if we look to the right of 56781020, we're going to get positive times positive times positive at 5 due to its multiplicity. The multiplicity on five is now 1, so it's going to turn negative at negative or at 4 plus root 6 / 2, we're going to turn positive. Because if we thought about factoring that, we know that we'd have two terms out of this quadratic here, so that each multiplicity would have to be 1. At 4 minus root 6 / 2, we're going to have to have its multiplicity change for the same reason. So now we should be able to figure out how the original function's going to go. It's going to be increasing but concave down. So if we're increasing but concave down, now we're going to be decreasing. So this is at 0, decreasing and down. So that's going to make this a local Max at 4 minus root 6 / 2. We're going to go increasing and we're going to be up. So we're going to, oh, I did that backwards. OK, first derivative here. So we're increasing, decreasing, decreasing, increasing, increasing, increasing second derivative. So we're down, we're down, we're up, we're up, we're down and we're up. OK, so let's sketch this again. Increasing bit, concave down, decreasing and down South local Max at 0. Now at 4 minus root 6 / 2, we're still decreasing, but now we're concave up, so that's a point of inflection. At 2:00, we're going to be increasing and concave up. So at two we're going to have a local min four plus root 6 / 2. We're still increasing, but now we're concave down, so this is going to be a point of inflection. At 5:00, we're increasing and we're concave up, so we're going to have another point of inflection. So that's a very rough sketch of what this graph would look like.