00:01
Okay, good day, ladies and gentlemen.
00:05
This is a problem with 357.
00:08
And i'm not going to write out the whole problem, but you can read it on your own if you prefer.
00:15
So we're starting with the function f of x equals to x 1 over x.
00:22
The first part is to figure out where the tangent line is horizontal.
00:28
So in particular then we're after we want to find a is to find x such that what does it mean to be horizontal? well, it means that f prime at x is equal to zero.
00:51
So let's first calculate.
00:54
Before we begin on that, let's actually calculate right.
01:00
So we have to calculate the derivative.
01:03
So how would we do this? well, what i want to do is i want to use the logarithmic differentiation.
01:14
So i have that y is equal to x to the 1 over x.
01:23
Okay, so then i get from here, from here i get, natural log of y i get that the natural log of y is equal to natural log x um or uh yeah natural log x maybe i do it this way yeah this way one over x times natural log x um okay so differentiating the left -hand side so now i take the derivative of both sides and i get the derivative this side here is me y prime over y so y prime divided by and now instead of writing y i'm just going to write this as x to the one over x okay so this is just again like i said this is just writing y and this is equal to now i'm differentiating the other side so this is equal to now the derivative of the other side of course this is a product so this is equal to the 1 over x prime times a natural log x plus 1 over x times um oops i want to do it 1 over x times natural log x prime okay so when we when we plug this in the derivative of 1 over x of course is negative 1 over x to the second and then multiplying that by natural log x plus, and now we have 1 over x times 1 over x.
03:48
And now i can, now what i want to do is simplify this, and i'm just going to use some algebra here.
03:59
And so what i end up with, of course, is oops.
04:05
This is equal to, let's see, 1 over x squared times 1 minus the natural log x.
04:22
So this is my derivative.
04:25
So this is equaled, i should say, multiplying by, if i multiply, maybe i do this, sorry.
04:34
I just want to move a little bit to make a little more room here.
04:38
Sorry about that, folks.
04:40
This is equal to that.
04:42
And then i can multiply by the x, 1 over x, to both sides.
04:48
And that tells me then, and so that tells me here that i get y prime is equal to x to the 1 .0 .1 .5.
05:05
Over x times 1 minus natural log x and then all of that of course divided by x divided by x squared so so this is my derivative now okay now the next question i have to consider is so i want i want to find the let's see so this is my derivative so now we want to find a so maybe i'll do it like this a is to find x such that such that f prime x is zero okay so that in other words that it's horizontal so when you look at this so notice that since x here has to be greater than zero and i didn't specify that i should have done so i apologize the domain here which is very important of course the domain is that x has to be greater than zero so the domain dom of f is equal to x greater than zero okay so that's actually very important in this whole problem you have to remember that the domain of f has to be greater than zero.
06:46
So when we look at the derivative here, you'll notice that x to the one over x divided by x squared, you know, this, this here, this part here can never be zero, okay? this is never zero, okay? so that means all we have to do is examine the one minus long x to get zero.
07:10
So let's do that here.
07:13
So we want to get 0 equals to 1 minus lon x.
07:25
And this, of course, means that we have that long x equals to 1.
07:36
The exponential of both sides we get, this is x equals to e.
07:44
So this is the point we want is that x equals e.
07:51
Okay, so this is the only point when the derivative is zero.
07:58
And now proceeding to the next step, so this breaks us into two intervals.
08:06
And one is the when the interval negative infinity to x.
08:17
Okay, this is our first interval...