 In this video, we're going to find all the relative extrema for three functions, and we're also going to determine on what intervals are these functions increasing or decreasing. So for our first example, we have the function f of x equals 2x cubed minus 3x squared minus 72x plus 15. So that we have this polynomial function. If we're looking for the local or relative extrema, we know that by Fermat's theorem these will occur at either end points if we have a closed interval, which we don't, or they'll happen at the critical numbers. The critical numbers are those numbers which make the first derivative go to zero or make the first derivative undefined. So the first thing we need to do is calculate the first derivative here. By the usual power rule, we're going to get that the first derivative is equal to 6x squared minus 6x minus 72. The derivative of a constant goes to zero, so it just disappears. And so this derivative itself is a polynomial and will be undefined anywhere. So we just need to figure out when is the first derivative equal to zero to help us find these critical numbers. So we have this quadratic equation we're trying to solve. We're going to do this by factoring if possible. If not, we could use the quadratic formula. We do notice that the coefficients 6, 6, and 72 are all divisible by 6. You can factor out the 6 leaving behind x squared minus x minus 12. Now we need to find factors of negative 12 that add up to be negative 1 for which we can do negative 4 and positive 3. So we factor this as x minus 4, x plus 3 equals zero. And so now we found two critical numbers. We have positive 4 and negative 3. If there were any domain restrictions, we would have to check right now to make sure that these numbers are inside the domain. But as it's a polynomial, the domain is all real numbers. No issue with that. So what we have to now do is test whether these critical numbers are local, extreme, or not. Just because you have a critical number doesn't mean it's necessarily a maximum or a minimum. We need to test it. And we're going to test this using the first derivative test. So what we're going to do is we're going to build a sign chart. You want to think of this line on the screen right now. It's like the x-axis. And we're going to mark it based upon these critical numbers. We have negative 3 and 4. So what we've done is we've taken the x-axis, the real line, and broke it up into three pieces. There's the interval negative infinity to negative 3. There's the interval negative 3 to 4. And then there's the interval 4 to infinity. And so we've broken up the x-axis in these three pieces. So what we want to do is now on these three intervals, we want to determine whether the derivative is positive or negative. So what can we say about the derivative here? So if you take, for example, the interval negative 3 to 4, how can we decide if the derivative is positive or negative on that? Now, one thing important to note here is that to build this sign chart over here we're using the intermediate value theorem. That is to say that because we know that the only places that the derivative is equal to 0 is at 4 and negative 3. These are the only critical numbers. We know by the intermediate value theorem because the derivative f prime is itself a continuous function. The only time it can change signs is at an x-intercept, aka these critical numbers. So if I pick any point, any number between negative 3 and 4, and I put that into the first derivative, that will be the sign for every number. So for example, what if I take the test point x equals 0? Well, we can actually do the derivative at 0. We end up with a negative 72. Notice that is, in fact, negative. And so we're going to record that the first derivative is negative if we're between negative 3 and 4. And it doesn't matter which test point you use. If you had instead done the first derivative at 1, we end up with 6 minus 6. Those just cancel out. It gives you a minus 72. That's likewise going to be 72 again. We could have done f prime at 2. That's a little bit more of a calculation we'd have to do there, but we would see that that's going to be less than 0. It's going to be negative. Again, by the intermediate value theorem, since the first derivative is continuous, any number between any value, any number actually picks between negative 3 and 4. If I put that into the first derivative, and we are putting it into the derivative here, it doesn't matter between negative 3 and 4. If you choose it, it'll be all of them are going to turn out to be negative here. What about something bigger than 4? Well, if you want to, you can take x to be 5. Whole numbers are good. We want to pick, if you're picking a test point, we want to pick one with easy arithmetic. 5 is pretty good. I also like 10 because it's a polynomial function. We have to take powers of 10. Powers of 10 are pretty easy. 10 square is just going to be 100 times e by 100 or 1000. Pretty easy. We can probably figure it out. Honestly, though, I like to do as little arithmetic as possible, so if I want to take a test value, it's a secret here. I'll use infinity because it's a lot easier to figure out what's going on here. Because with a polynomial function, if you take the limit as x approaches infinity, then only the leading term matters. What happens to the quadratic as x approaches infinity? Well, what happens to 6x squared? 6x squared will approach infinity. In particular, it's a positive infinity, so we see that this side here is going to be positive. Then by similar reasoning, if we take the limit as x approaches negative infinity of the derivative, we get 6x squared. It's a positive leading coefficient, has an even degree, so as x approaches negative infinity, y will still approach positive infinity, so we get this, plus minus plus. Filling out this sign chart is exactly what we do to help us determine what's going on here. But I also want to mention another approach that you can use in the situation that we don't just have to use these test points. One strategy that some people like to use is to use factorization here. If we just consider the factors of the derivative, x minus 4 and x plus 3, we found these factors because, well, we needed them to find the critical numbers, but if you look at them just as by themselves, it's just a linear function, y equals x minus 4, this is a linear function with a positive one slope. Therefore, this function's beginning going to be increasing as you span the x-axis. It'll start on the negative side, it'll end with the positive side. When does it switch from negative positive because a line won't only do it once, it'll switch at its x intercept, which is in this case 4. So it's going to be below the x-axis until you hit 4 and then it'll be positive afterwards. So notice if I have this linear factor, I can fill in this sign chart without any calculations whatsoever, negative, negative plus. I don't need to do any arithmetic there. And then if you take the line y equals x plus 3, the other factor in the derivative, then again, it's going to start off negative. It'll switch to pause when you hit its x intercept, which is negative 3. So you get negative positive positive, in which case then the first derivative is the product of these two things. I mean, there's also a product of 6, mind you, but as it's positive 6, that's not going to contribute much. And so notice what happens here. When you're less than negative 3, you're going to have a negative times a negative, that's going to be a positive. When you're between negative 3 and 4, you get negative times positive, which is a negative. And when you're greater than 4, you get a positive times a positive, which is equal to a positive. So if you use the factorization of the derivative, then you don't have to use any of these test points, which the test points aren't so hideous, but it avoids a little bit of arithmetic, which honestly is the hardest part of calculus. So that's its alternative strategy that I will suggest to you. We'll see this on the subsequent examples in just a moment. So however you want to fill out this sign chart, one third approach I'll mention, as we're trying to graph a parabola, we know the basic graphs can look something like this. It has an x intercept at negative 3, has an x intercept at 4, it's a concave up parabola because the leading coefficient is positive. So we know the graph of this parabola is going to look something like this, so it's going to be it's going to be positive, negative, positive. So again, there's a couple ways you could try to fill in this sign chart. Now that we've now that we've identified the signs of the derivative, we then can infer from the signs of the derivative the monotonicity of the original function f, because but whenever the derivative is positive, that means that the function is going to be increasing. Whenever the derivative, first derivative is negative, that means the function will be decreasing. And again, when it's when the first derivative is positive, that means the function's increasing like so. So we basically see that our functions can be doing something like the following. It's going to be going up, then it's going to be going down, and then it's going to be going down, and then it's going to be going up. So what does this tell us about our critical numbers, negative 3 and negative 4? We see that negative 3 is going to turn out to be a local maximum and we see that 4 is going to turn out to be a local minimum. So let's summarize the information that we've discovered so far here. So what we know is the following. We see that f is increasing, it's increasing on the interval negative infinity to negative 3 union 4 to infinity. We see that f is decreasing, it's decreasing on the interval negative 3 to 4. We have, so f has a local max at the value x equals negative 3 and it has a local min at the value x equals 4. So we all determine all of this information using the first derivative that is the first derivative test. Let's look at another example. This time we don't have a polynomial function but it is still a linear combination of power functions so it'll be very similar in nature. Let's figure out where the local extrema of this function, where is it increasing, where is it decreasing. So we'll start off with the first derivative. By the power rule, we're going to get the first derivative f prime is 6 times 2 thirds times x to the negative 1 third. That's a 1 right there and then you're going to subtract from that 4. Now this one we have to be a little bit more careful about because we do want it to see when it's equal to 0 but we also are interested in what makes it undefined. Because you have a negative x one in here, that actually does mean division and so we can see there are going to be some values that make this thing undefined. So let's try to write the derivative as a fraction next. Some things to note, 3 goes into 6 of course 2 times, so I'll leave you with 2, 2 times 2 gives you a 4. We can factor out a 4 since that's a common coefficient. I'm also going to factor out x to the negative 1 third power. What does that leave behind? Well we took away the 4, we took away the x to the negative 1 third so it gives us a 1 right there and then we subtract from the negative 4 that we factor out the 4 and then we also divide it by x to the 1 third. So if you divide by a negative it actually gives you something positive and so rewriting it we see that the first derivative looks like 4 times 1 minus x to the 1 third power. That's just the cube root there and this sits above x to the 1 third power. So what makes the denominator go to 0? That's an important thing that we have to identify here because if the denominator goes to 0 that makes the expression undefined. You have x to the 1 third equals 0 if you take the third, if you cube both sides take the third power. You're going to get x equals 0 and so this is going to be a critical number and we can actually see that this is going to correspond to a vertical tangent line, a vertical tangent. We're also interested in what makes the numerator go to 0 because if the numerator goes to 0 that actually makes the whole fraction go to 0 and that's going to coincide with a horizontal tangent line. So we get 4 times 1 minus x to the 1 third power equals 0. Divide both sides by 4. We get 1 minus x to the 1 third power equals 0. We'll just move the cube root to the other side so we get the x to the cube root. That is the 1 third power equals 1. If we cube both sides we're going to get x equals 1 right here and this is going to coincide with a horizontal tangent line. So recording what we've discovered here we have two critical numbers. We have x equals 0 and x equals 1 as our critical numbers. So now that we have our critical numbers 0 and 1 let's investigate the monotonicity of the graph. We're going to build a sign chart based upon our first derivative so we're going to leave that on the screen as we draw this thing. So we have 0 and we have 1. So let's consider the factors of f prime here. So one of the factors we can ignore the 4 because there's just a constant positive 4 there. One of the factors will take 1 minus the cube root of x here like so and the other factor we're going to take y to be the cube root of x like so. So can we figure out what's going to go on with this function right here? Let's take the cube root of x for example. This is a graph we know very well. The graph of this thing looks kind of like the following. This function is always going to be increasing but it'll switch from negative to positive once you hit its x-intercept which will happen at 0. So we see that our function here is going to go from negative it'll be positive positive for the remaining intervals. What about the function y equals 1 minus x the cube root? Well in that case if we think about in terms of transformations the negative sign here what that does is it's going to reflect the graph so now the cube root is going to look something like this and then the plus 1 shifts everything up by one changing the x-intercept but this is what we need to see here it's going to go from positive to negative negative there. So in particular this thing will look like positive positive negative like so and so if we put this together the first derivative at prime we see that when you look at the combinations a positive and negative gives you a negative a positive a positive gives you a positive and a negative and a negative excuse me a negative and a positive gives you a negative. So this gives us the signs of the first derivative for which then we can infer the monotonicity of f we see that f is going to be decreasing then increasing then decreasing on its domain. So if we summarize what we've seen here so it's the f is decreasing it's decreasing on the interval negative infinity to zero union one to infinity so notice I got the information so I want everything to the left of zero that's negative infinity to zero I want everything to the right of one that's going to be one to infinity there and we know that the function is increasing it's increasing on the interval zero to one what about its extrema f has what can we say about these critical numbers well if the function was decreasing but then becomes increasing that's an example of a local minimum by the first derivative test and if it was increasing then decreasing that means our function was a has a maximum at x equals one so let's say this here f has a local minimum at the value zero x equals zero and it has a local maximum at x equals one like so now we were able to figure out this fill out this signed chart using factorization of the derivative if you wanted to you could just use test points for example you could pick something larger than one something between zero and one and then something less than zero you could do that but merely there's no whole numbers between zero and one so you know if you're gonna if you're gonna use test points I'd probably do something like let's perfect q's would be good so let's say like x equals eight if you're above x equals negative one in this interval you could do like x equals one eighth uh you know that's that's that's a perfect cube fraction so you could use that one to help you fill out the chart if you prefer using the arithmetic this arithmetic approach of test points like we did on the previous example let's finish this video with one more example here let's find the intervals of monotonicity for f of x this time to equal x times e to the two minus x squared we do have to calculate the first derivative the product rule is going to come into play this time we take the derivative of x which is one then we get a e to the two minus x squared next we're going to get x times the derivative of e to the two minus x squared which when you take the derivative of natural exponential you get back itself but the chain rule comes into play and we take the inner derivative which is going to be negative two x like so and so this is our derivative we want to set this equal to zero notice that when you look at this function it's just a bunch of power functions and exponentials there's no place where it's going to be undefined so we just have to figure out when it's equal to zero so we need to factor this thing the first thing to recognize is that both of these have these powers of e both of the terms there so let's factor out the power of e so we're going to get e to the two minus x squared that leaves behind a one plus x times negative two x right there for which keeping the exponential still there we get one minus two x squared this is equal to zero now one great thing about exponential functions is that exponentials if you take e to any power whatever you want most called theta this can never equal zero in fact these exponentials are always positive so there's no way that the exponential factor could equal zero it must in fact be the case that one minus two x squared equals zero which if we add two x squared of both sides we get one equals two x squared divide by two we're going to get x squared equals one half and therefore our critical numbers are twofold we get that x equals plus or minus the square root of one half which if you prefer you could write that as plus or minus one over the square root of two or plus or minus the square root of two over two these are all the same number it really kind of comes down to if you like rationalizing denominators are such if you want to approximate as they are irrational you get plus or minus zero point seven oh seven like so and so we're going to build our sign chart based upon this idea right here let's draw that one so we're going to mention our critical numbers so we have negative the square root of one half or negative point zero seven if you prefer and then we're going to positive the square root of one half like so and we want to think about the first derivative if you're going to approach it using factorization notice that e to the two minus x squared is always positive so it's never going to change the sign if you're using factors it would look like one minus the square root of one half x and then y equals one plus the square root of one half x you can do that one but again I don't really want to use irrational numbers if I don't have to we could use a test point like you use x equals zero it's over here we can take something bigger than the square root of one half they could be like x equals one x equals negative one negative one and put that into the first derivative and calculate those things I'm going to use the fact that the essentially it looks like a parabola and you know there is a factor of the exponential but as that's always positive I only care about the signs it looks like it's an upward parabola excuse me it's a concave down parabola because of the negative coefficient there so I can see that the first derivative the sign chart is going to look like negative positive negative which then what does that mean about the function f it means that we were decreasing then we were increasing and then we were decreasing so in particular when you approach the negative square root of one half you go down and then up so that suggests we have a minimum and then for the square root of one half you are increasing the decreasing so that means you have a maximum so let's then report then what we found for this function here so f is increasing it's increasing on the interval we're going to take negative square root of one half to positive the square root of one half and it is decreasing on the interval this time we're going to get negative infinity all the way up to negative the square root of one half union we get positive square root of one half up to infinity that's where it's decreasing we see that f has a local minimum at the value x equals negative square root of one half and we see that it has a local maximum at the value x equals the square root of one half and so we've been demonstrated in this video how we can find the local extrema of a function using the first derivative test