Absolute maxima and minima Determine the location and value of...

Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = 3x⁵ - 25x³ + 60x on [-2,3]

Accepted Answer

Welcome back, everyone. On the interval open bracket -23 close bracket, find the absolute maximum and minimum of the function F of X equals 2 X 5 minus 20 X cubed plus 50 X. Now if we're going to find absolute extreme values, we'll need to find the critical points, then we'll need to test if they are maximum and minimum and if we have more than one maximum maximum or more than one minimum, then we can test which one is the absolute values. So let's start by first finding our critical points, OK? And we do that by setting the derivative of FX to 0. So at the critical points. The first derivative will be equal to 0 and know if we differentiate F of X, then that means we're going to get 10 x 4 minus 60 X2 plus 50 to be equal to 0. Now we need to solve for X. Notice that this looks a bit like a quadratic equation. So if we let Y be equal to X2, OK, then we could rewrite this as 10 Y2 minus 60 Y plus 50 equals 0. No, as a quadratic equation, we can tell that A is 10, B is -60, and C is 50. So by the quadratic formula. OK, we can solve for Y and then solve for X. So by the quadratic formula, that means Y will be equal to negative B 60. Plus R minus the square root of B2. Minus 4 multiplied by a multiplied by C. I know all of this is being divided by 2 multiplied by A. Now, we can go ahead here and solve for Y. Let me come over to the right, OK. And then we should find that we get Y to be equal to 60 plus or minus 40, OK. 60 plus or minus 40 divided by 20. Now what does that mean? Well, if we add, that means Y will be equal to 100 divided by 20, which is 5. And if we subtract, OK, so let me write that up here. So plus or minus means it's 60 + 40 divided by 20 or 60 minus 40 divided by 20. And when we subtract, then we get Y to be equal to 1. Remember, Y equals X2. So this means that X2 equals 5 or X2 E equals 1. And in that case, that means X equals plus R minus root 5 or it equals plus or minus 1, OK? Because the square root of 1 is plus or minus 1. Now, here. If we think about it, when we have our values for X, we know that the negative value of route 5 would lie outside the given interval. Remember our interval is from -2 to 3 inclusive. So since uh negative route 5 lies outside the interval. OK. Then our possible values for X are route 5. Or plus or minus 1. Now, what does that mean? These are critical, these are the X values for critical points, but what does that mean? Well, what we're saying here is that if we were to put them on the X axis, OK, then here we could say that we have our points. So starting at -2. OK. Then we're gonna have -1. It's gonna go from 0 to 1, and then we're gonna have route 5, and then we're gonna have 3, OK? So, here we have our critical points and our endpoints, and no, the absolute maximum or minimum could be any one of these points. So we can test which one it is by using the first derivative test. And essentially we're going to look at the sign, OK, of our first derivative before and after these points, OK? And if it goes from If it goes from negative to positive, it's a local minimum, but if it goes from positive to negative, it's a local maximum. So let's test each of those with the first derivative test, OK? Uh testing their nature. Or testing each point with the first derivative test. OK. They know we're gonna take our points. So first, let's start with the interval from -2 to 1, OK? So far -2 up to -1, OK. Then we can use a value of X, let's say X equals -3, 1.15. When that's the case, then our first derivative. It's gonna be equal to 10 multiplied by -3 to the 4th, minus 60 multiplied by -3 to the second. Plus 50 Which when we evaluate that we get it to be -275 divided by 8, so in this interval. X or sorry, our first derivative is negative. I'll just write that on the axis that I have here. Next, for the interval between our root, well, let's see here, between -1 and 1. OK. Let's use X equals 0. And when X equals 0, then our first derivative is gonna be equal to 10 multiplied by 0 to 4th. OK. Minus 60 multiplied by 0 to the 4th plus 50, which equals 50. Thus it's going to be positive in this interval. Next, OK, let's go to when. X is in the interval from 1 to route 5. We write that properly here. And now in this interval, let's use a value for X to be equal to 2, OK? When X equals 2, the first derivative is going to be equal to 10 multiplied by 2 to the 4th minus oops, 60 multiplied by 02. So I just realized here, this should have been 0 squared in what I wrote previously, OK. So 660 multiplied by 2 squared, my apologies, friends, I was just fixing here. So that's 60 multiplied by 22 plus 50, which is gonna be equal to -30. So in that interval, it's negative. Finally, for the interval from route 5 to 3, let's let X be equal to 5 halves or 2.5. Then the first derivative is going to be equal to 10 multiplied by 5/2 to the 4th minus 60 multiplied by 5/2 plus 50. Which equals 525 divided by 8. So in this case, it's positive for that interval. No, what can we use to help us evaluate which ones are maximum or maximum or minimum, well. Notice that at X equals -1, OK, at X equals -1. The sign changes from negative to positive. So this implies a local minimum, OK? So I could probably write it. Let me make some space down here, I think. And let me just carry this as well, just to so that we have something to look on. OK. So now, notice that here, what is this telling us first that if. Of -1, so it says X equals -1, sorry, is a minimum, the local minimum. Next, notice that when X E equals 1, the sign changes from positive to negative, so X E equals 1 is a local maximum. And finally At route 5, OK, the sign changes from negative to positive, so X equals route 5 is also a local minimum. But remember in our problem we want to find our absolute maximum and minimum values. Well, here we know that X equals 1 is going to be the absolute maximum because it's the only maximum from our list. But what about -1 and rule 5? Which one of these are the absolute minimums? Well, let's just substitute our values here, OK? So now to figure it out, we can find F of -1 and F of root 5 to determine which is the absolute, and then we'll also find F1 to determine the actual value of the absolute maximum. So for F of -1, that's going to be 2 multiplied by -1 to the 5th, minus 20 multiplied by -1 cube plus 50 multiplied by -1, and that equals -32. Now, to find the rest of values, I'll leave that up to you, but it's pretty much doing the same thing, replacing X4 or a value. So for F of 1, it's eventually going to be equal to 32, OK. Uh, when X equals route 5, then it's gonna be equal to 0. And if we also take our end points. When X equals 2, then sorry, when X equals -2. OK. Then F of -2 equals -4, and when X equals 3, OK. Then F of 3 equals 96. So what does this tell us? Well, no, when we compare these values we can find. That the absolute minimum. Is when X equals -1, which is 32, and the absolute maximum, OK, the absolute maximum, let me write that properly here. Absolute minimum. Equals -32 at X equals -1. And the absolute maximum. I 96 at X equals 3, OK? So these are the absolute minimum and maximum values for our equation within the given interval. Thanks a lot for watching everyone. I hope this video helped.

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.ƒ(x) = 3x⁵ - 25x³ + 60x on [-2,3]

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Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = 3x⁵ - 25x³ + 60x on [-2,3]