Consider the function
http://prntscr.com/90izfw

- amy0799

Consider the function
http://prntscr.com/90izfw

- Stacey Warren - Expert brainly.com

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- jamiebookeater

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- TrojanPoem

Take the derivative for the function:
f'(x) = 2 + 3cos(x)
let , f'(x) = 0
3cos(x) = -2
cos(x) = -2/3 (negative in 2nd, 3rd quarter) ( in the interval (0, 360) )
x = 180 - 48.18 = 131.8 , x = 180 + 48.18 = 228.18
|dw:1447011573858:dw|

- amy0799

increasing: (infinity, 131.8) and (228.18, infinity)
decreasing: (131.8, 228.18)
?

- TrojanPoem

No

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## More answers

- TrojanPoem

The interval given is in (0, 360) so :|dw:1447012031542:dw|
I've a little lag.

- amy0799

i dont understand what you mean

- TrojanPoem

"Consider the function below in the interval ( 0, 2pi)"
That means, our end is 2pi , start 0.
I am using numbers for easier calculations, but you should use pi , 2pi, ... instead.

- amy0799

how do i figure that out?

- TrojanPoem

To turn an angle from degrees to radians multiply times pi/180

- TrojanPoem

360 * pi/180 = 2pi
228 * pi/ 180 = 19pi/15 (round to the nearest)

- amy0799

how did you get 48.18?

- TrojanPoem

shift cos 2/3 on the calculator

- amy0799

cos(2/3) = 0.79

- TrojanPoem

Shift + cos

- TrojanPoem

You will get the cosine inverse cos^-1

- amy0799

arccos(2/3)=.84

- TrojanPoem

48.1896851

- amy0799

that's not what i got on my calculator

- TrojanPoem

hhhhhh xD

- TrojanPoem

what do you do in your calculator ?

- amy0799

arccos(2/3)=0.84107

- amy0799

ooh, i had mine in radians

- TrojanPoem

Ok, at least , I knew the cause of this.

- amy0799

isn't it easier to have in radians so you dont have to convert it from degrees to radians?

- TrojanPoem

Sometimes, but in our case if you got the angle in degrees and rounded it you can put it in a better form like 228 degree = 19pi/15

- amy0799

i'm gonna have to put my answer in decimals, so isn't it more accurate to be in radians first?

- TrojanPoem

It is.

- amy0799

so then do i add and subtract 180 to 0.84?

- TrojanPoem

You 're using radians so 180 = 3.14 = pi so subtract 3.14 - 0.84

- amy0799

3.14-180=2.30
3.14+180=3.98

- TrojanPoem

3.14 - 180 ?

- amy0799

oops its suppose to be 0.84

- amy0799

oh, how do i fix this?

- TrojanPoem

It's right, I am just used to degree :o

- amy0799

oh ok, how do i find where it's increasing and decreasing?

- TrojanPoem

If you 've an interval (a,b) put the a at the start point of the number line, b at its end. now find your critical points f'(x) = 0, put them on the number line get a number between them if it's > 0 then type +++ , <0 ---- , after , before, done.

- amy0799

|dw:1447014429594:dw|

- TrojanPoem

|dw:1447014529785:dw|

- amy0799

isnt the critical point -2/3?

- TrojanPoem

cosx = -2/3
x = cos^-1(-2/3)

- TrojanPoem

we are plotting the numbers line for the x variable not cos

- amy0799

|dw:1447014782121:dw|

- TrojanPoem

Good, but you can get the sign if it's increasing or decreasing from the derivative , like take 3 , put x =3 ( middle) , x = 5 (sign of after), before(2.30) take x = 1

- amy0799

so plug 3 into f'(x)?

- TrojanPoem

yep

- amy0799

|dw:1447015049266:dw|

- TrojanPoem

nice

- TrojanPoem

+ = increasing , - = decreasing

- amy0799

increasing: (infinity,2.30) and (3.98, infinity)
decreasing: (2.30, 3.98)
?

- TrojanPoem

increasing [0, 2.3[ and ]3.98, 2pi]
decreasing ]2.3, 3.98[
at critical points the function is constant, so not increasing or decreasing , so it will be an open interval at the critical point.

- amy0799

shouldn't it be in braces?

- TrojanPoem

] << for intervals [1,2] 1 , 1.1 , 1..00001 , 1.999999 , 2 (all numbers between)
{ << for group of numbers {1,2} only 1, 2
() << for pairs maybe coordinate,

- amy0799

in my attachment, it's in braces

- TrojanPoem

It works too.

- amy0799

ok. how do i find the relative minimum and relative maximum extrema?

- TrojanPoem

|dw:1447015726005:dw|

- TrojanPoem

2.3 is max , 3.98 is min

- TrojanPoem

if the function was increasing then it started decreasing so you have a maxima, if the function was decreasing and started increasing so it's a minima

- amy0799

ok, i got it correct. can you help me with another problem?

- TrojanPoem

Sure, but did you understand ? I think my explanation is pellet xD

- amy0799

yea, i understand it, it just when u had it in degrees was when i got the most confused

- amy0799

http://prntscr.com/90k9k1
i already plugged in the values for C(t), i think they are correct

- TrojanPoem

You want me to re-check the numbers ?

- amy0799

yea and how to do the problem please

- TrojanPoem

Not quiet sure about the first part, but in physics, we get the average (2+2.5)/2 = 2.25
The second part, graph it using any online graphing tool and approximate it
the last one get the first derivative , put your signs and get the maxima

- TrojanPoem

c(T) = 3t(26+t^3)^-1
c'(t) = 3(26+t^3)^-1 - 3t(26+t^3)^-2 * 3t^2
let c'(t) = 0
3 (26 + t^3) - 3t * 3t^2 = 0
78 + 3t^3 - 9t^2 = 0
-6 t^3 = -78
t^3 = 13
t = cubicroot(13)
t = 2.35 hour

- amy0799

i got everything right except on the chart part for C(t) =1, and 2

- TrojanPoem

I didn't review your numbers.

- amy0799

sorry, i was checking my work. Thank you. I have one my problem i need help with

- TrojanPoem

Sure, but hurry as I am going to do the workout.

- amy0799

http://prntscr.com/90kj9k

- TrojanPoem

veclorty is the change of displacement over the time.
v = ds/dt

- amy0799

so i plug in 0 and 60 into V(t)?

- TrojanPoem

Get the derivative

- amy0799

12t-8

- TrojanPoem

find where the function is increasing or decreasing

- amy0799

how?

- TrojanPoem

Like we did in the 2 last problems.

- TrojanPoem

v(t) = 12t - 8
12t = 8
t = 2/3
|dw:1447017827218:dw|

- TrojanPoem

can you induce where the particle is moving in positive direction, negative, changing direction ?

- amy0799

positive: (2/3, 60)
negative: (0, 2/3)

- TrojanPoem

|dw:1447018062402:dw|

- TrojanPoem

Change direction ?

- amy0799

not sure how to figure that out

- TrojanPoem

At 2/3 (the critical point) at this point the particle wasn't either in positive or negative direction.

- amy0799

thank you very much, and sorry for taking up your time!!

- TrojanPoem

No problem :)

- TrojanPoem

Good luck, I wish you had understood.

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