00:06
Okay, so today we're going to be working on the problem of finding the critical points of the function capital f of x, which is equal to x to the power of four -fiths, multiplied by x minus four.
00:46
So recall first what the critical points of a function are.
00:52
These are exactly the values of x for which the derivative of f is equal to zero.
01:02
So we're searching for the values of x, we can plug into the derivative, which will make it equal to zero.
01:08
And just as a reminder, recall to that this means something geometrically.
01:13
So if we, now, this graph i'm going to draw doesn't necessarily represent this function f, but if i have some sort of graph of a function looking like this, remember that our critical points, these are the places which are the local maximum or a local minimum.
01:33
So it may not be the highest or lowest point on the entire graph, but within a small region it will definitely be the lowest point.
01:45
Okay.
01:46
So what we need to do then is we need to find the derivative of our function.
01:51
If we're going to find the critical points.
01:54
That would be step one.
01:56
Now, usually in this case, well, here we only have a quadratic term.
02:03
And so i think in this case, it would actually be pretty easy to just expand this out and use our rules of differentiation for monomials.
02:13
And i would actually invite you to try that method on your own.
02:19
But i see here a good opportunity for us to practice a couple different rules of differentiation.
02:25
So why don't we take that opportunity? so i want to figure out the derivative of this function.
02:33
And clearly, just looking at it, oh, it's not obvious what the derivative of this looks like.
02:38
So maybe i can use some of my rules of differentiation to break this down into more understandable chunks.
02:47
Well, this first piece here, x to the four -fifths, that's just a monomial in x, and we have a very simple rule for differentiating that.
02:58
So this is something i do know how to differentiate just by looking at it.
03:03
The other piece is a bit more complicated, but if i could at least pull this bit off, then i know i can solve part of it.
03:12
So if we think about, well, here's the piece i know how to solve.
03:16
What's its relation to the rest of the function? well, it's being multiplied.
03:21
In other words, there's a product.
03:25
So that's our clue then that, well, maybe the product rule is a good thing to apply right away here.
03:31
So why don't we go ahead and do that? so the derivative of f at x using the product rule, i'm going to get that this is the derivative of x to the four -fifths multiplied by x -minus -4 squared plus x -to -the -four -fifths.
04:00
Sorry, i forgot to put a prime there, a little tick indicating that we're taking the derivative.
04:07
So it's going to be x to the four -fifths, and we'll be taking the derivative of x -minus 4 -squared.
04:22
So right away, we know what the first derivative is over here.
04:27
That's why we decided to use the product rule because we saw, oh, here's a piece of this function.
04:32
I do know, i do understand how to differentiate.
04:34
Let's go after that first.
04:36
This other part here, though, this seems a little bit more tricky.
04:40
Now, again, this is something we could expand out and differentiate pretty easily.
04:45
But why don't we, this is a good chance for us to practice using.
04:50
The chain rule here.
04:52
So if you just cover up that exponent 2, we see that we have the function just x minus 4, and that's again something we can just differentiate.
05:04
We have very simple rules to differentiate that.
05:08
And then if you covered up the x minus 4 and just imagine that as an x, well, it would be x squared.
05:13
And again, that's a very easy function to differentiate.
05:16
So the key here is, well, i've got two functions i know how to differentiate, and one is inside of the other, right? the x minus 4 is being substituted into the x squared.
05:30
So here we can use the chain rule.
05:33
So in this case, if i let g of x be x squared and i let h of x be x minus 4, sorry, i didn't mean to make that mark there.
06:01
Well, then we can see that the function we're interested in differentiating is just going to be g composed with f.
06:18
So we're going to have x minus 4 squared.
06:27
The derivative of this is equal to the function g evaluated at the function h.
06:38
Well, we'll be the derivative of this whole thing.
06:41
And the chain rule tells us that the derivative is the derivative of g evaluated at h, multiplied by the derivative of h.
07:00
So what we're going to need to do here then is we need to figure out what the derivative of g is and what the derivative of h is.
07:09
Well, the reason why we picked this decomposition is because these are very easy derivatives to solve.
07:16
We just use a rule for monomials on g to get that the derivative is 2x.
07:21
And we use our rule for monomials on h to get that the derivative is just one.
07:30
So, first we take g prime, which is 2x, and we replace each instance of x with the function h of x.
07:43
So this is just going to become 2 multiplied by x minus 4.
07:50
And h prime in this case is just one.
07:54
And so here we have the derivative of the second term.
08:00
So the derivative of this term here is 2 times x minus.
08:16
So evaluating this derivative then, just move to a new page here.
08:26
So we have f prime of x is equal to.
08:32
Now, here we have x to the four -fiths.
08:36
Well, its derivative, just using the rule for monomials, is going to be four over five times x to the four over five minus one.
08:49
But i have fractions here.
08:53
So if i'm going to do addition or subtraction, i need the denominators to be the same.
08:58
So i'm going to write one in a sort of tricky way.
09:01
I'm going to write one is 5 divided by 5.
09:06
5 divided by 5 is 1, but i'm just rewriting it in this different way so that the denominators are the same, and i can in fact subtract these numbers now...