Answer:
Step-by-step explanation:
f(t) = t + cos(t)
Criticals :
f' (t) = 1 - sin (t) = 0
sin (t) = 1
Given that the interval is : [ - 2π , 2π ]
thus; t = π/2 and -3π/2
The region then splits into [ -2π , -3π/2 ], [-3π/2 , 2π ] and [ π/2 , 2π ]
Region 1: Region 2: Region 3:
[ -2π , -3π/2 ] [-3π/2 , 2π ] [ π/2 , 2π ]
Test value (t): -11 π/6 Test value(t) = 0 Test value = π
f'(t) = 1 - sin(t) f'(t) = 1 - sin(t) f'(t) = 1 - sin(t)
f'(t) = 1 - sin (-11 π/6) f'(t) = 1 - sin (0) f'(t) = 1 - sin(π)
f'(t) = positive value f'(t) = 1 - 0 f'(t) = 1 - 0.0548
thus; it is said to be f'(t) = 1 (positive) f'(t) = 0.9452 (positive)
increasing. so it is increasing so it is increasing
So interval of increase is : [ -2π , 2π ]
There is no local maximum value or minimum value since the function increases monotonically over [ -2π , 2π ]. Hence, there is no change in the pattern.
c) Inflection Points;
Given that :
f'(t) = 1 - sin (t)
Then f''(t) = - cos (t) = 0
within [ -2π , 2π ], there exists 4 values of t for which costs = 0
These are:
[-3π/2 ]
[-π/2 ]
[π/2 ]
[3π/2 ]
For Concativity:
This splits the region into [ -2π , -3π/2], [ -3π/2 , -π/2], [-π/2 , π/2] , [π/2 , 3π/2] and [ 3π/2 , 2π].
Region 1: [ -2π , -3π/2] Region 2: [ -3π/2 , -π/2]
Test value = - 11π/6 Test value = π
f''(t) = - cos (t) f''(t) = - cos (t)
- cos (- 11π/6) = negative f''(-π) = - cos (- π)
Thus; concave is down. f''(π) = -cos (π)
positive, thus concave is up
Region (3): [-π/2 , π/2]
Concave is down
Region (4): [π/2 , 3π/2]
Concave is up
Region (5): [ 3π/2 , 2π]
Concave is down
We conclude that:
Concave up are at region 2 and 4: [ -3π/2 , -π/2] , [π/2 , 3π/2]
Concave down are at region 1,3 and 5 : [ -2π , -3π/2] , [-π/2 , π/2] , [ 3π/2 , 2π]
