00:01
We're going to find and classify the stationary points of the following functions.
00:05
First one is y equals negative x squared plus x plus one.
00:10
So the first derivative is negative 2x plus one.
00:14
And the second derivative is negative 2.
00:25
So we're going to set the first derivative negative 2x plus 1 equal to 0.
00:29
Subtract 1 from both sides.
00:31
We're going to solve for x here.
00:33
I get negative 2x equals negative 1.
00:35
Divide both sides by negative 2, and i get x equals one half.
00:41
This is my x value of either my minimum or my maximum.
00:45
I'm going to go ahead and take this x value and evaluate it in my first original function.
00:51
Y equals negative 1 1⁄2 squared plus 1⁄2, and we get y equals 1 .25.
01:01
So at 1⁄2 .5, we have a maximum.
01:07
And i know this is a maximum because my leading coefficient here is negative, which means this is going to be a and it's squared.
01:18
So that exponent means it's a parabola.
01:20
It's a u -shaped graph.
01:22
But that negative flips it upside down.
01:25
So it's going to look something like this.
01:27
So i'm going to have a maximum.
01:29
It's not going to be at zero like that.
01:31
But it would be at one half, one point two five.
01:33
So let's say it's there.
01:34
And it would look something like this.
01:37
Y equals x squared minus 4x plus 4.
01:41
My first derivative is 2x minus 4 and my second derivative is 2.
01:48
2x minus 4 equals 0.
01:51
Add 4 to both sides.
01:53
2x equals 4, divide both sides by 2.
01:57
And we find our x value is 2.
01:59
That's the x value of either our maximum or our minimum.
02:03
This is actually going to be a minimum.
02:06
And we can know that even before we find the y value, based off what our function here tells us, it's squared.
02:14
So it's a u -shaped graph.
02:15
It's positive...