Finding Max and Mins Part IV

[Hckyplayer8]

Determine the absolute max and min values of f(x) = x - 2 sin x on the interval [0,2pi] and the x values where they occur.

f'(x) = 1-2 cos x
f(0) = 0
f(2pi) = 2pi - 2

Setting the derivative to is 1 - 2 cos x = 0 results in x = 1/2. Using the unit circle, cos 1/2 can equal pi/3 or 5pi/3

Inserting those values back into the original function.

f(pi/3) = pi/3 - 2 sin pi/3 = pi/3 - 2 (radical 3 over 2)

Does that look correct thus far?

 

[Harry_the_cat]

I suggest you calculate f(2pi) again. The rest looks good so far.

 

[Hckyplayer8]

I suggest you calculate f(2pi) again. The rest looks good so far.

Ah.

Sin (2pi ) is 0.

2 x 0 = 0.

2pi - 0 = 2pi.

Thanks!

 

[Hckyplayer8]

Continuing on

f(pi/3) = pi/3 - 2 sin pi/3 = pi/3 - 2 (radical 3 over 2) = pi/3 - radical 3

I believe that is all I can do, no?

Making sure I get my terminology correct here...f(5pi/3) ends up 5pi/3 - (-radical 3) because it lies in quadrant IV which is just a reflection of quadrant I of the unit circle?

If that is all correct, then how can I tell the max and mins of this function? Visualizing where the above two points would be on a graph is not coming to me.

 

[Harry_the_cat]

Remember that x takes the values between 0 and 2pi inclusively. So your x-axis goes from 0 to 2pi.
You've calculated 4 different y values correctly so now compare them (you may have to turn them into decimal form to be able to compare them) in the usual way to find the abs max and min values.

 

[Deleted member 4993]

Continuing on

f(pi/3) = pi/3 - 2 sin pi/3 = pi/3 - 2 (radical 3 over 2) = pi/3 - radical 3

I believe that is all I can do, no?

Making sure I get my terminology correct here...f(5pi/3) ends up 5pi/3 - (-radical 3) because it lies in quadrant IV which is just a reflection of quadrant I of the unit circle?

If that is all correct, then how can I tell the max and mins of this function? Visualizing where the above two points would be on a graph is not coming to me.

Have you learned to check f"(x) [i.e. is the second derivative f(x)] for max or min?

 

[Hckyplayer8]

Remember that x takes the values between 0 and 2pi inclusively. So your x-axis goes from 0 to 2pi.
You've calculated 4 different y values correctly so now compare them (you may have to turn them into decimal form to be able to compare them) in the usual way to find the abs max and min values.

The decimal trick worked...although I needed a math solver to get it done. Are there any tricks to learning those conversions so I will be able to do it come test time?

 

[Hckyplayer8]

Have you learned to check f"(x) [i.e. is the second derivative f(x)] for max or min?

Not in my class. Although it appears you fine folks on here hooked me up in another thread.

 

[Harry_the_cat]

It's handy to know that sqrt(2) = 1.414 approx., sqrt(3) =1.732 approx. and of course pi = 3.1416 approx.

 

[Deleted member 4993]

Not in my class. Although it appears you fine folks on here hooked me up in another thread.

In that case, follow Cat's advice at post #7 above.