Question

How do you get the abs. extrema for f(x)=cos(piX) on the closed interval [0, 1/6]?

Answer 1

Hello there ms kateryna! :)

I'm not really familiar with `absolute` extrema.
I guess we need to first determine that the function is continuous on the given interval, yes?
Which......... uhh, it's cosine, with a period of 2 so it's continuous. If we need to do more than that for our justification you can let me know.

So for extrema we'll take the derivative of f(x) and then set the function equal to zero.
Have you tried taking the derivative yet?\[\Large f(x)\quad=\quad \cos(\pi x), \qquad\qquad f'(x)\quad=\quad ?\]

Answer 2

yeah, I got f'(x)=-pisin(pix)

Answer 3

dont know where to go from there....

Answer 4

Ok good! :)
Setting our derivative function equal to zero:\[\Large f'(x)\quad=\quad -\pi \sin(\pi x),\qquad\qquad 0\quad=\quad-\pi \sin (\pi x)\]And from here we'll try to solve for x.

Answer 5

Dividing each side by -pi gives us,\[\Large 0\quad=\quad \sin (\pi x)\]

Answer 6

the pi in the sin(pix) stays even though you divide it out?

Answer 7

We divided out the pi that was in front.
The pi inside is stuck inside! We can't mess with it.

Answer 8

ok

Answer 9

From here, we need to remember our special angles.\[\Large 0\quad=\quad \sin(\color{royalblue}{\theta})\qquad\to\qquad \color{royalblue}{\theta\quad=\quad?}\]When does sine give us 0? :) Which angles?

Answer 10

180 and 0?

Answer 11

Good good. Let's write them in radians not degrees though.
So the first 2 angles would be \(\Large 0\) and \(\Large \pi\).
\[\Large 0\quad=\quad \sin(\color{royalblue}{\theta})\qquad\to\qquad \color{royalblue}{\theta\quad=\quad0,\;\pi}\]

So with our problem:\[\Large 0\quad=\quad \sin (\color{royalblue}{\pi x})\qquad\to\qquad \color{royalblue}{\pi x\quad=\quad 0, \; \pi}\]
Does that make sense kind of?
Our `angle` is the thing inside of the sine function.
So we're relating our special angle to pi*x.

Answer 12

got it

Answer 13

So we found a couple solutions for pi*x,\[\Large \pi x\quad=\quad 0, \qquad\qquad \pi x\quad=\quad \pi\]Let's divide through by pi,\[\Large x\quad=\quad 0, \qquad\qquad x\quad=\quad 1\]So these are a couple of critical points we were able to find.
If we spun around the circle we would be able to find others as well.
But we were constricted to a small domain of [0 , 1/6].
Do both of these critical points fall in that region?

Answer 14

no, only the 0 falls in the range

Answer 15

what does that mean?

Answer 16

So that means we can't use the critical point x=1, it isn't within our interval. So we don't care about that point.

So we are able to conclude that there is only 1 critical point within our interval, and it happens to be at x=0.

Do they want us to do anything more for this problem? Like classify it as a max/min point?

Answer 17

yep

Answer 18

i also need to graph it

Answer 19

Ok to find our what type of point we have, let's run a couple of test points through our derivative function.

We want to find out the `sign` the derivative produces on the left side and right side of our critical point.\[\Large f'(x)\quad=\quad -\pi \sin(\pi x)\]So let's pick a nice convenient number on the left side of x=0.

How aboutttttt, x=-1/6,
That will give us,\[\Large f'(-1/6)\quad=\quad -\pi \sin(-\pi/6)\]

Answer 20

What does sin(-pi/6) give you? :x

Answer 21

-1/2? or 45*

Answer 22

ya -1/2 sounds right,\[\Large f'(-1/6)\quad=\quad (-\pi)(-1/2)\]So this result is `positive`, yes?
That is telling us that the function is `increasing` on the left side of our point x=0.

Answer 23

Let's check the right side of x=0 somewhere (within our interval).
So how bout at x=1/6,\[\Large f'(1/6)\quad=\quad -\pi \sin(\pi/6)\]

Answer 24

negative

Answer 25

Ok good good :)
Understand what I was doing with the arrows?
Based on the arrows, can you determine if that's a max or min?

Answer 26

yeah, relative maximum

Answer 27

Ok good! :)
We would also want to check the `end points` of our interval to compare and see if those values give us something interesting.

x=0 happens to be our critical point, so we've already dealt with that end point.
Then ummm, since we had no other critical points, I guess we can call x=1/6 a relative minimum since x=0 was a max and x=1/6 would have to be smaller than that.

Answer 28

ok, i can do the rest like the box graph and line graph, etc.

Answer 29

Got it from there? Cool!

Oh oh I forgot, we should list our max/min as ordered pairs.
So be sure to plug your x=0 and x=1/6 back into f(x) and write them as ordered pairs. :)

Answer 30

ok, thank you so much for taking the time to explain everything to me instead of just giving me the answers, you have been really helpful