MU Tutorial Sheet Finding Relative Extreme Points (Application differentiation) ❤️🫶

Added: 2026-05-10

[00:00:00]So, in this particular lesson, we are still looking at application of differentiation.

[00:00:07]We are going to be looking at what it means if you have been asked to find a relative extrema of a given function.

[00:00:15]Okay? And extrema, we are simply trying to talk about the relative uh local minimum point of a particular function or the relative uh uh local maximum point of a particular function.

[00:00:34]So, when we say extrema, extreme extrema, we are simply trying to see where the curve is increasing, where it is decreasing. Okay? That's uh really something that you must be able to to be able to know. Now, it depends with the instruction that you have been given by the question. You can be asked to find the relative extrema of a function using the second derivative. Of course, we use the second derivative because the second derivative is what is going to tell you whether the the uh the the nature of the root of the points that you're going to find, whether your curve is going to be increasing or it will be decreasing.

[00:01:14]It's very important.

[00:01:16]Now, I'm going to show you very simple steps that you're going to be using whenever you want to find extrema of a given function.

[00:01:24]Let's start.

[00:01:26]And of course, if you want to register for personal lessons, make sure you inbox me on this WhatsApp number, plus 26077664046.

[00:01:40]Okay? That's the number that you need to contact me on.

[00:01:44]Let's begin.

[00:01:47]We find the relative extrema of the following function using the second derivative.

[00:01:52]The function that we have been given is f of x equal to x to the three minus three x to the two minus 24 x plus 32. Let's find the first derivative.

[00:02:09]And this implies that f prime or dy dx is equal to This is going to be three x to the two.

[00:02:17]Three x to the two minus six x minus 24.

[00:02:23]From there, we need to find the critical points. So, from the first derivative, find the critical points.

[00:02:32]So, the critical points are going to be found at some point where f prime is equal to zero.

[00:02:46]Then from there, we can just divide everything by what is common.

[00:02:52]So, this is going to be zero is equal to x to the power two minus two x minus Okay, what is three into 24? This kind more like eight or something.

[00:03:09]Okay, let me just Okay, so 24 divide by three eight. I want to make a mistake.

[00:03:19]Okay, then from there, we can find the product the sum. Product is equal to negative eight, the sum is equal to negative two. The factors are going to be Which number if you multiply Okay, negative four and two. Let's see.

[00:03:37]So, if we multiply negative four by two, we'll be getting negative eight. If we add negative four plus two, we'll be getting negative two. Okay, cool. So, from there, we are going to have x to the power 2 plus 2x - 4x - 8 = 0 then x is common x + 2 -4 is common x - x or x + 2 Okay, so this holds.

[00:04:10]So our critical points are found at some point x + 2 then x - 4 = to some zero.

[00:04:18]Then x is going to equal to -2 then x is going to equal to 4. So the first derivative should be able to give you critical points.

[00:04:30]The first derivative should be able to give you critical points.

[00:04:36]We have found the critical points.

[00:04:38]Now we get the second derivative.

[00:04:42]Now dy over dx2 We now get the equation the second the first derivative which is this.

[00:04:59]Then we find dy dx again.

[00:05:04]This is going to give you 6x -6 Now when you reach here in the first lesson I had already shown you that whenever your d power 2x over dx to the power 2 is less than zero your curve is going to be like this.

[00:05:27]Then it implies that you have a maximum value here.

[00:05:32]But if your d over dx2 is greater than zero your curve is going to be like this then it implies that you have a minimum point here.

[00:05:45]So the critical points that you found are what you shall be using to test.

[00:05:50]So, let's try to test.

[00:05:54]This is supposed to be d to the power 2 y over dx to the power 2. So, now let's test what it is.

[00:06:02]I'm going to put I'm going to get 6, then I'm going to put negative 2 here.

[00:06:11]Notice that I'm going I'm I'm getting negative 12 minus 6.

[00:06:15]And this is giving me like minus 18.

[00:06:19]Notice that d 2 y over dx2 is equal to negative 18.

[00:06:27]Negative 18 is less than zero, right?

[00:06:32]Negative 18 is less than zero. So, this implies that at x is equal to negative 2, we are having a relative maximum point.

[00:06:49]We're having a relative maximum point.

[00:06:51]So, the maximum point is going to be at negative 2 comma then you get the original equation.

[00:07:07]You put your negative 2 here.

[00:07:11]Negative 2 here.

[00:07:14]It's going to be long.

[00:07:17]Negative 2 here plus 32.

[00:07:26]Let me just write everything.

[00:07:28]Otherwise, this is actually very simple.

[00:07:33]You get your y is equal to negative 2 to the power 3 minus 3 negative 24 - 2 + 3 2. Now, let's try to check what are we going to get.

[00:07:51]So, - 2 to the power 3 - 3 - 2 to the power 2 - 24 then - 2 + 32.

[00:08:19]So, this is giving us Y is equal to 60.

[00:08:23]So, at the maximum point is is is at some point -2, 60. Simple.

[00:08:31]That's how you solve this.

[00:08:33]Now, let's try to check the second one.

[00:08:36]The second one we can try 4. We see what 4 is going to give us. What are we using to test?

[00:08:43]We're using the same thing to test, the second derivative, which is this equation. So, this is going to be d squared of y over dx to the 2 is equal to where there is x I'm going to put a 4.

[00:09:03]And this is going to be like 6 multiplied by this is going to be 24 - 6. Let me see.

[00:09:10]24 - 6 Aha! Getting 18.

[00:09:15]So, you get your 18, right?

[00:09:17]Now, look.

[00:09:22]Your d power 2 of y over over dx to the 2 this time around is greater than 0. When it is greater than 0 you're having a minimum point here. So, the minimum point is at some point 4, then let's try to put our four into the equation.

[00:10:02]You get the critical point, you put it directly in the equation, you see what you're going to get. So, this time around we have y is equal to Let's find the y value, so it's going to be 4 power 3 then 3 4 to the power 2 24 3 plus 32.

[00:10:21]Let's see what we are going to get.

[00:10:25]So, open bracket 3 Sorry, open bracket 4 to the power 3 minus 3 4 there power 2 minus 24 open bracket 3 plus 32.

[00:10:50]Sorry.

[00:10:54]So, 4 to the power 3 minus 3 4 to the power 2 minus 24 open bracket 4 inside plus 32.

[00:11:10]Okay, I get you 48.

[00:11:12]So, this whole thing is y equal to 48.

[00:11:15]So, at 4, 48 at 4, -48 we are having the minimum value. So, this is the maximum extrema point, that's the minimum relative extrema point. That's how you will be able to solve this. Actually, very simple. Let's do the second one just for the sake of clarity and assurance that we have understood what we are doing and we are not guessing.

[00:11:51]So, we have the function of f of x is equal to x to the 2 minus 16 over x.

[00:11:59]Right? So, you can find dy dx of this.

[00:12:03]Then, of course, we know that f prime, which is dy dx, is equal to 2x minus then this is going to be minus 16 over x to the 2.

[00:12:16]So, f prime is equal to 2x plus 16 over x to the 2.

[00:12:25]We make f prime equal to zero.

[00:12:29]We get the critical points.

[00:12:34]Goes this side, it becomes negative. 2x is equal to plus 16 over x to the 2. Want to get the critical points, we cross-multiply.

[00:12:45]We are going to get minus 2x to the 3 is equal to 16.

[00:12:52]Minus two, minus two.

[00:12:55]x is equal to x cubed is equal to negative three.

[00:12:59]We find the cube root of negative eight.

[00:13:03]And your x is equal to negative two.

[00:13:09]Okay, so you have a point there, x, the critical point, x is equal to negative two.

[00:13:15]I think this one should even be very easy. And we'll find the second derivative.

[00:13:23]The second derivative, we are just finding dy dx of the first.

[00:13:28]So, the first was this.

[00:13:35]So, the second derivative we find dy/dx of this going to be 2 - the same 16 but this time around to the three.

[00:13:45]We shall be using this to be able to test whether it's going to be a minimum or maximum point.

[00:13:51]How do we do that?

[00:13:53]We're going to get a two minus the 16 over We get the critical point.

[00:14:04]What is uh negative two to the power three? It's going to be 16 over negative eight.

[00:14:12]So, it's going to be two.

[00:14:16]This and this will cancel. This is going to be like plus two.

[00:14:26]Okay?

[00:14:27]Then this will be like a four. Let me just try to confirm.

[00:14:30]2 minus 16 all over by minus eight.

[00:14:38]Okay, yeah, it's a four.

[00:14:40]Now, notice that four Notice that when you you you get the minimum point at some point dy/dx to the power two when it is less than zero, that's when you get the maximum point rather.

[00:14:55]The curve is going to be in this way.

[00:14:59]The maximum point is found at this point.

[00:15:02]But if you have dy/dx to the two greater than zero, you have a minimum point at this point, right?

[00:15:11]Now, because our in this case we have found four, so meaning d2 over dx2 is actually greater than zero. So, we have a minimum point. So, we have a minimum point at -2, {comma} we get whatever we have, we put it in the original equation.

[00:15:38]Okay? Get what we have, we put it in the original equation.

[00:15:42]So, the original equation was -2 ^ 2 which is this one -16 over -2.

[00:15:54]So, this is going to be like 44.

[00:15:58]Plus, this is going to be like eight.

[00:16:02]So, Y is equal to 12.

[00:16:07]You have a minimum point at that.

[00:16:10]You can do C.

[00:16:12]I can start for you.

[00:16:15]So, in case maybe you want to part them up, you want to separate them.

[00:16:19]Of course, your C is just X over Okay, over over one.

[00:16:28]Plus, X over X to the power two.

[00:16:35]Let me see.

[00:16:39]If it will be able to hold. So, this won't be able to hold. So, that basically means you'll be using quotient to be able to find dy dx of that.

[00:16:50]Or you can use your chain rule where you're going to say X then open bracket one plus X to the two to the power -1. Use your chain rule.

[00:17:03]Here, rather product rule. Okay? Product chain rule.

[00:17:06]And then you'll be able to find your things. But, this is how you are able to find the relative maximum extrema point, relative minimum extra extra Thank you so much.