amy0799
  • amy0799
Consider the function http://prntscr.com/90izfw
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TrojanPoem
  • 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
  • amy0799
increasing: (infinity, 131.8) and (228.18, infinity) decreasing: (131.8, 228.18) ?
TrojanPoem
  • TrojanPoem
No

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TrojanPoem
  • TrojanPoem
The interval given is in (0, 360) so :|dw:1447012031542:dw| I've a little lag.
amy0799
  • amy0799
i dont understand what you mean
TrojanPoem
  • 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
  • amy0799
how do i figure that out?
TrojanPoem
  • TrojanPoem
To turn an angle from degrees to radians multiply times pi/180
TrojanPoem
  • TrojanPoem
360 * pi/180 = 2pi 228 * pi/ 180 = 19pi/15 (round to the nearest)
amy0799
  • amy0799
how did you get 48.18?
TrojanPoem
  • TrojanPoem
shift cos 2/3 on the calculator
amy0799
  • amy0799
cos(2/3) = 0.79
TrojanPoem
  • TrojanPoem
Shift + cos
TrojanPoem
  • TrojanPoem
You will get the cosine inverse cos^-1
amy0799
  • amy0799
arccos(2/3)=.84
TrojanPoem
  • TrojanPoem
48.1896851
amy0799
  • amy0799
that's not what i got on my calculator
TrojanPoem
  • TrojanPoem
hhhhhh xD
TrojanPoem
  • TrojanPoem
what do you do in your calculator ?
amy0799
  • amy0799
arccos(2/3)=0.84107
amy0799
  • amy0799
ooh, i had mine in radians
TrojanPoem
  • TrojanPoem
Ok, at least , I knew the cause of this.
amy0799
  • amy0799
isn't it easier to have in radians so you dont have to convert it from degrees to radians?
TrojanPoem
  • 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
  • amy0799
i'm gonna have to put my answer in decimals, so isn't it more accurate to be in radians first?
TrojanPoem
  • TrojanPoem
It is.
amy0799
  • amy0799
so then do i add and subtract 180 to 0.84?
TrojanPoem
  • TrojanPoem
You 're using radians so 180 = 3.14 = pi so subtract 3.14 - 0.84
amy0799
  • amy0799
3.14-180=2.30 3.14+180=3.98
TrojanPoem
  • TrojanPoem
3.14 - 180 ?
amy0799
  • amy0799
oops its suppose to be 0.84
amy0799
  • amy0799
oh, how do i fix this?
TrojanPoem
  • TrojanPoem
It's right, I am just used to degree :o
amy0799
  • amy0799
oh ok, how do i find where it's increasing and decreasing?
TrojanPoem
  • 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
  • amy0799
|dw:1447014429594:dw|
TrojanPoem
  • TrojanPoem
|dw:1447014529785:dw|
amy0799
  • amy0799
isnt the critical point -2/3?
TrojanPoem
  • TrojanPoem
cosx = -2/3 x = cos^-1(-2/3)
TrojanPoem
  • TrojanPoem
we are plotting the numbers line for the x variable not cos
amy0799
  • amy0799
|dw:1447014782121:dw|
TrojanPoem
  • 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
  • amy0799
so plug 3 into f'(x)?
TrojanPoem
  • TrojanPoem
yep
amy0799
  • amy0799
|dw:1447015049266:dw|
TrojanPoem
  • TrojanPoem
nice
TrojanPoem
  • TrojanPoem
+ = increasing , - = decreasing
amy0799
  • amy0799
increasing: (infinity,2.30) and (3.98, infinity) decreasing: (2.30, 3.98) ?
TrojanPoem
  • 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
  • amy0799
shouldn't it be in braces?
TrojanPoem
  • 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
  • amy0799
in my attachment, it's in braces
TrojanPoem
  • TrojanPoem
It works too.
amy0799
  • amy0799
ok. how do i find the relative minimum and relative maximum extrema?
TrojanPoem
  • TrojanPoem
|dw:1447015726005:dw|
TrojanPoem
  • TrojanPoem
2.3 is max , 3.98 is min
TrojanPoem
  • 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
  • amy0799
ok, i got it correct. can you help me with another problem?
TrojanPoem
  • TrojanPoem
Sure, but did you understand ? I think my explanation is pellet xD
amy0799
  • amy0799
yea, i understand it, it just when u had it in degrees was when i got the most confused
amy0799
  • amy0799
http://prntscr.com/90k9k1 i already plugged in the values for C(t), i think they are correct
TrojanPoem
  • TrojanPoem
You want me to re-check the numbers ?
amy0799
  • amy0799
yea and how to do the problem please
TrojanPoem
  • 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
  • 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
  • amy0799
i got everything right except on the chart part for C(t) =1, and 2
TrojanPoem
  • TrojanPoem
I didn't review your numbers.
amy0799
  • amy0799
sorry, i was checking my work. Thank you. I have one my problem i need help with
TrojanPoem
  • TrojanPoem
Sure, but hurry as I am going to do the workout.
amy0799
  • amy0799
http://prntscr.com/90kj9k
TrojanPoem
  • TrojanPoem
veclorty is the change of displacement over the time. v = ds/dt
amy0799
  • amy0799
so i plug in 0 and 60 into V(t)?
TrojanPoem
  • TrojanPoem
Get the derivative
amy0799
  • amy0799
12t-8
TrojanPoem
  • TrojanPoem
find where the function is increasing or decreasing
amy0799
  • amy0799
how?
TrojanPoem
  • TrojanPoem
Like we did in the 2 last problems.
TrojanPoem
  • TrojanPoem
v(t) = 12t - 8 12t = 8 t = 2/3 |dw:1447017827218:dw|
TrojanPoem
  • TrojanPoem
can you induce where the particle is moving in positive direction, negative, changing direction ?
amy0799
  • amy0799
positive: (2/3, 60) negative: (0, 2/3)
TrojanPoem
  • TrojanPoem
|dw:1447018062402:dw|
TrojanPoem
  • TrojanPoem
Change direction ?
amy0799
  • amy0799
not sure how to figure that out
TrojanPoem
  • TrojanPoem
At 2/3 (the critical point) at this point the particle wasn't either in positive or negative direction.
amy0799
  • amy0799
thank you very much, and sorry for taking up your time!!
TrojanPoem
  • TrojanPoem
No problem :)
TrojanPoem
  • TrojanPoem
Good luck, I wish you had understood.

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