I don\'t understand this one, please help. f(t)=t+cost, -2Pi leq t leq 2Pi (a) F

Question

I don't understand this one, please help. f(t)=t+cost, -2Pi leq t leq 2Pi (a) Find the interval of increase. (Enter your answer using interval notation.) (-infinite ,infinite) x (b) Find the local maximum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) DNE Find the local minimum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) DNE (c) Find the inflection points. (x, y) = ( ) (smallest x-value) (x, y) = ( ) (largest x-value) Find the intervals the function is concave up. (Enter your answer using interval notation.) Find the intervals the function is concave down, (Enter your answer using interval notation.)

Explanation / Answer

f(t) = t + cos(t)

Deriving :

f'(t) = 1 - sin(t) = 0

sin(t) = 1

t = -3pi/2 and pi/2

So, this splits the domain into (-2pi , -3pi/2) , (-3pi/2 , pi/2) and (pi2/ , 2pi)

Region 1 : (-2pi , -3pi/2)
Testvalue = -11pi/6
f'(t) = 1 - sin(t)
f'(-11pi/6) = 1 - sin(-11pi/6)
f'(-11pi/6) = 1 + sin(11pi/6) ---> positive value
So, INCREASES

Region 2 : (-3pi/2 , pi/2)
Testvalue = 0
f'(t) = 1 - sin(t)
f'(0) = 1 - sin(0)
f'(0) = 1 --> positive
So, INCREASES

Region 3 : (pi/2 , 2pi)
Testvalue = pi
f'(pi) = 1 - sin(pi)
f'(pi) = 1 --> positive
So, INCREASES

Therefore, intervalof increase : (-2pi , 2pi) ----> ANSWER

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Inflection :

f''(t) = -cos(t) = 0

cos(t) = 0 over -2pi <= t <= 2pi

So, t = -3pi/2 , -pi/2 , pi/2 and 3pi/2

When t = -3pi/2 , y = -3pi/2 + cos(-3pi/2) --> y = -3pi/2
When t = -pi/2 , y = -pi/2
When t = pi/2 , y = pi/2
When t = 3pi/2 , y = 3pi/2

So, inflection points are :

(-3pi/2 , -3pi/2) ---> ANSWER
(-pi/2 , -pi/2) --> ANSWER
(pi/2 , pi/2) --> ANSWER
(3pi/2 , 3pi/2) --> ANSWER

Concavity :

Regions of interest are (-2pi , -3pi/2) , (-3pi/2 , -pi/2) , (-pi/2 , pi/2) , (pi/2 , 3pi/2) and (3pi2/ , 2pi)

Region (-2pi , -3pi/2)
Testvalue = -11pi/6
f''(t) = -cos(t)
f''(-11pi/6) = -cos(-11pi/6) = -cos(11pi/6) --> negative
So, CONCAVE DOWN

Region (-3pi/2 , -pi/2)
Testvalue = -pi
f''(-pi) = -cos(-pi) = -cos(pi) = 1 --> positive
CONCAVE UP

Region (-pi/2 , pi/2)
Testvalue = 0
f''(0) = -cos(0) = -1 --> negative
CONCAVE DOWN

Region (pi/2 , 3pi/2)
Testvalue = pi
f''(pi) = -cos(pi) = 1 --> positive
CONCAVE UP

Region (3pi/2 , 2pi)
Testvalue = 11pi/6
f''(11pi/6) = -cos(11pi/6) --> negative
CONCAVE DOWN

So, concave up : (-3pi/2 , -pi/2) U (pi/2 , 3pi/2) --> ANSWER
Concave down : (-2pi , -3pi/2) U (-pi/2 , pi/2) U (3pi/2 , 2pi) --> ANSWER

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