Calc. Min/Max problems (extrema of y = x - cosx, etc)

paulxzt

Junior Member
Joined
Aug 30, 2006
Messages
65
Please help me with these. I sort of solved them but I dont think I did it right because the answer choices seem different

1. Find all extreme values in the interval [0,2pi] for y = x - cosx

I found the derivative to be y' = 1 + sin x. Then I set it equal to zero and got sin x = -1, x = 3pi/2. What do I do from here? Do I find the value of f'(3pi/2) and f'(0) ? ::The answer choices are given like (-1,3pi/2) , (pi , 1+pi ) , (-1 , 0 ) , (3pi/2 , 2pi)::

2. Find the minimum value of the slope of the curve y = x^5 + x^3 - 2x

I found y' = 5x^4 + 3x^2 - 2

Now what do I do?

3. The motion of a particle is given by position function s(t) = t^3 - 6t^2 + 12t - 8
The minimum value of the speed is

Do I just find the derivative and then set it equal to 0? and then plug it in to determine which is the minimum?

I appreciate any help. Thank you!
 
paulxzt said:
Please help me with these. I sort of solved them but I dont think I did it right because the answer choices seem different

1. Find all extreme values in the interval [0,2pi] for y = x - cosx

I found the derivative to be y' = 1 + sin x. Then I set it equal to zero and got sin x = -1, x = 3pi/2. What do I do from here? Do I find the value of f'(3pi/2) and f'(0) ? ::The answer choices are given like (-1,3pi/2) , (pi , 1+pi ) , (-1 , 0 ) , (3pi/2 , 2pi)::

See which one falls in your interval?.

xcosxoa3.jpg


[quote:3n44l35x]2. Find the minimum value of the slope of the curve y = x^5 + x^3 - 2x

I found y' = 5x^4 + 3x^2 - 2

Now what do I do?
[/quote:3n44l35x]

Find the 2nd derivative, set to 0 and solve for x. Sub that value into the 1st derivative and that will be the minimum slope.

\(\displaystyle \L\\y''=20x^{3}+6x\)

\(\displaystyle \L\\20x^{3}+6x=0\), x=0

\(\displaystyle \L\\5(0)^{4}+3(0)^{2}-2\)

See it?.
 
Top