Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Increasing and decreasing functions. Let F be a function defined on an interval I and let X1 and X2 be two points in I. If F of X1 less than F of X2, whenever X1 is less than X2 then F is increasing on I. Or we could think of it as if the first derivative is positive or the slopes are positive at each point on an interval, then F is increasing on that interval. If F of X1 is greater than F of X2 whenever X1 less than X2, then F is decreasing on the interval I. Or if the first derivative is less than 0 at each point on an interval, then F is decreasing. So if the slope is always negative, then we're decreasing on that interval. The word monotonic means only increasing or only decreasing on some interval I. If we look at some example. Oh, not yet the first derivative test for local extrema. Suppose that C is a critical point of a continuous function F and that F is differentiable at every point on some interval containing C except possibly at C itself. Moving across the interval from left to right, If F prime changes from negative to positive at C, then F has a local minimum at C. So if our slope goes from a negative slope to a positive slope, if we think about a negative slope to a positive slope, then at that Point C we had to have a local minimum. If F changes from a positive slope to a negative slope, then at that C we have to have a local Max. And if F prime does not change signs at C, that is, if it's positive on both sides of C or negative on both sides of C, then it has no local extrema at sea. So it's going to be positive here, but on the other side it's also still being positive. It's going to be called a point of inflection for this kind of occurrence. It's going to be undefined for a cusp or a corner like before. This one's going to also be a point of inflection, so we did get a horizontal slope of 0 there. But the original function, the first derivative, didn't change from positive to negative on those two locations. So let's look at some examples. If we're given the first derivative and we're asked to figure out the critical points and where it's increasing or decreasing the local Max and min. So if we're given this first derivative, we could think of this as X -, 3 / X to the half equaling the first derivative or X - 3 / sqrt X. So we know that we can only look at 0 to Infinity because the inside of a square root would have to be positive and we can't have zero in the denominator. So critical points, we're going to let this equal 0 and we're going to solve. So we're going to have 0 equal X -, 3 or X equal 3. We know that we have a critical point at our end point of 0 also. So if we look at this, if we think about 0 to three, what's happening for our first derivative if we're to the right of three, say 4, if I stick in four here, our first derivative is positive so that we know our original function here is increasing in between zero and three. If I put in two, say, I'm going to get negative Y values for that first derivative. So we know that our original function here has got to be decreasing. So this location here is going to be a local min at X equal 3, and this one's going to be a local. Oh, it's not going to be a local Max because we don't include the point. So there is no local Max because it wasn't included. So our original function F is increasing from three to Infinity, and our original function F is decreasing from zero to three. Let's look at some more examples. This one, the directions are going to change. We're going to be asked where it's increasing and decreasing, and then we're going to be asked to find the local and absolute extrema. So the first thing we're going to do is find our first derivative. And our first derivative is going to be 6 -, 3 Theta squared. We're going to set that equal to 0. So we'd get Theta squared equaling 2, or theta's going to equal the positive and negative sqrt 2. If we thought about putting this just on a number line to figure out what's happening to the right of root 210 a hundred a million. If I have 6 - 3 * 10 ^2, it's a negative number. Now we could actually think of our multiplicity doing this. Also, if we thought about taking this and factoring out of three, we'd get 2 minus Theta squared, and then that could equal 32 minus the square root of Theta and two plus the square. Oops, sqrt 2, sqrt 2 minus Theta, sqrt 2 plus Theta. So we could see that those multiplicities are each one. So it's going to be negative to positive to negative. If you didn't see that, we could test points all the way through. So if we put in zero in this first derivative, we'd get a positive first derivative value. We took something to the left of negative root 2, we'd get a negative. So we're going to have increasing for the original function from negative root 2 to root 2. We're going to have decreasing on the original function from negative Infinity to negative root 2, union root 2 to Infinity. Now if we're decreasing, we know that our original function's going down and then we're going to go increasing. So our original function's going up. So we know that we have a local min at X equal negative root 2, and if we stick in negative root 2 to the original, we get 6 times negative root 2 minus negative root 2 ^3 or -6 root twos +2 root twos for -4 root 2. So local min at X equal -2 of -4 root 2. Here if we have a positive function, we know we're increasing and then it's turning to a negative function. We know that that's going to be a local Max at X equal root 2. If we stick that in, we get +4 root 2 for our Y value because the original function is a cubic and because the leading coefficient is a negative. We know that our graph is going to look something like this rough, rough, rough idea, and if that's the case, we're not going to have any absolute Max or mins, so no absolute extrema. Looking at another example, same directions this time. So we're going to have t ^3 -, 3 T squared. If we start with our first derivative, we're going to have three t ^2 - 6 T Setting that equal to 0, I'm going to factor out A3 TT -2 so we can see t = 0 and two with both multiplicities being one. We're also given an end point this time, so we're going to have 320 if we're in between 2:00 and 3:00, say 2 1/2. If I stick in 2 1/2 here, 3 * 2 1/2 ^2 - 6 * 2 1/2, that's going to give me a positive value in here. So in between zero and two, two had a multiplicity of 1. So we've got to change from positive on one side to negative, and 0's got a multiplicity of 1. So we're going to change again. It might have been easier to choose one to stick in here. We would have had 3 -, 6, which is a negative value, and then we could have done the other two looking at multiplicities at the zero and two. So we know we're going to be increasing from negative Infinity to 0 union two to three. We're going to be decreasing from zero to two. We're going to have F OF0F of two and F of three. So F of 0 is 0F of two is -4, and F of three is 0. So this graph would look like if we had a 0 here and a three here, we'd be doing something like this. So we know that this is an absolute Max and another absolute Max and a local min. I'm looking at another example going back to here. Remember if the first derivatives positive, we're increasing, first derivatives negative we're decreasing. So that had to be a Max negative to positive. So that had to be a min positive. So that had to be a Max. And then it took a matter of looking and thinking through whether they were going to be absolutes or not. It's AT cubed equation with a positive coefficient, so we know it had to do something like that if it kept going, but we were looking at an interval from negative Infinity to three. An example with trig functions, we're given an interval from negative π halves to Pi halves. So when we take the first derivative, we're going to get 2 secant X by bringing down the power. Then we have to take the derivative of that secant X by using the chain rule. Remember, think of this original as secant X quantity squared. So we brought down the two to one less power times the derivative of the inside, and then the derivative of tangent was secant squared. So when we take the derivative of the secant X, we're going to get secant tangent or secant times secant as secant squared tangent. When we set this equal to 0, we're going to factor out what they have in common, which is A2 secant squared X, leaving us tangent X - 1. So secant squared X = 0 and tangent X - 1 = 0. Well, this never happens. Secant's always negative Infinity to -1 or one to Infinity tangent is one at π force and five Pi force. Or we could think of it as π force plus K π. But we have to look at the interval between negative π halves and Pi halves. So we really are only looking at when X is π force. We're not including the end points. So all we have to look at is F of π force. And if we put an F of π force to the original, the secant of π force squared -2 tangent of π force, well, secant of π force is root 2 and when we square that tangent of π force is 1. So we actually are going to get 0. So at π force and we're on a closed, we're on an open interval, not closed because we didn't include the endpoints. So in between π force and Pi halves, if we look at the first derivative, if I chose actually let's choose zero, so let's go in between negative π halves and π force. At 0 secant of 0 is one tangent of 0 is 0. So 2 * 1 ^2 * 0 - 1 would give me a negative value. So this is going to be negative here. The Pi force came from this term here, which has an odd multiplicity, so it's got to be positive on the other side. So our original function is decreasing on negative π halves to π force, and our original function is increasing on π force to Pi halves. And we're going to have a local min here. Remember, if it's decreasing, the original function's going down and then increasing it's going up the original function, that actually should be an absolute min. Because if we're decreasing on this interval and we're increasing on this interval, that's got to be an absolute min. And we're not defined at the end point, so we don't have any maxes. Thank you and have a wonderful day.