00:01
We have the function f of x equals pi over 7 times x plus cosine of pi times x over 7 over the interval from x is greater than or equal to 0 and less than or equal to 14.
00:11
We want to find the absolute minimum of f of x on that interval and the absolute maximum of f of x on that interval.
00:17
So, i'll write our function again down here.
00:21
It is pi times x divided by 7 plus the cosine of pi x over 7.
00:29
And that interval is x is greater than or equal to 0 and less than or equal to 14.
00:35
So, to find the absolute maximum and the absolute minimum, we would need to find the derivative first.
00:41
So, f prime of x.
00:42
The derivative of pi times x over 7 would just be pi over 7.
00:49
And then the derivative of cosine, that would be a negative sign.
00:54
And then the derivative of cosine of whatever we're taking the cosine of, that would be pi times x over 7.
01:02
So, pi x over 7.
01:05
And then we need to multiply by the derivative of pi x over 7, which would just be pi over 7.
01:12
So, we can simplify that a bit.
01:14
We can just rearrange some factors.
01:16
That would be pi over 7 minus pi over 7 times the sine of pi x over 7.
01:22
Okay.
01:24
So, the absolute minimum or maximum would occur at the critical points of this possibly.
01:33
Or at an end point of the interval.
01:34
So, we need to find the critical point.
01:36
So, we would set the derivative equal to 0.
01:38
And we could, if we rearrange this equation a bit, we would have pi over 7 times the sine of pi x over 7 equals pi over 7.
01:52
So, then we divide both sides by pi over 7.
02:00
And the pi over 7 is cancelled.
02:02
And we're left with the sine of pi x over 7 equals, this would just give us 1.
02:09
So, then we would need to solve this equation.
02:13
So, if we take the inverse sine of both sides, or if we just look at this as an angle.
02:21
The sine of angle theta equals 1 at what? so, what angle has a sine of 1? so, that would be every pi over 2.
02:35
And because we're thinking on the unit circle.
02:37
On our unit circle there.
02:41
That's the point.
02:44
That's the point 0, 1.
02:46
And remember that the sine of theta would be the y coordinate there.
02:52
And that would be an angle of pi over 2.
02:57
So, any multiple of pi over 2.
03:02
We could just continue adding 2 pi.
03:06
So, pi over 2 plus 2 pi n.
03:10
So, pi over 2 plus any full rotation or n could be any integer.
03:15
So, that tells us that pi over 7, pi times x over 7 must be equal to pi over 2 plus 2 pi n.
03:27
So, then if we solve for x.
03:30
We would multiply both sides by 7 over pi.
03:36
Multiply both sides by 7 over pi.
03:40
And this would cancel and we're left with x equals.
03:43
If we multiply pi over 2 times 7 over pi.
03:46
That would just give us 7 over 2 plus 7 over pi times 2 pi n.
03:52
That would give us 14.
03:55
And the pi would cancel and we would just be left with 14 n.
03:58
So, x is 7 over 2 plus 14.
04:05
And that would be the times when the function has critical points.
04:10
Or when the derivative is equal to 0.
04:13
So, if we look in the interval.
04:19
So, we'll just let n equal 0 first of all.
04:22
If n equals 0 that means x would be 7 over 2.
04:25
And keep in mind we need to be in the interval from 0 less than or equal to x less than or equal to 14.
04:30
So, this value is in that interval.
04:33
So, what about if n equals 1? that means x would be 7 over 2 plus 14.
04:40
Which would be.
04:41
That would be past that interval.
04:45
Because that is not in the interval that's greater than 14.
04:50
So, we would ignore that...