Consider the equation below. f(x) = 3 sin x + 3 cos x, 0 LE x LE 2 pi Find the i
ID: 2868333 • Letter: C
Question
Consider the equation below. f(x) = 3 sin x + 3 cos x, 0 LE x LE 2 pi Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) Find the local minimum and maximum values of f. local minimum value local maximum value Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.)Explanation / Answer
f(x) = 3sinx + 3cosx
Deriving :
f'(x) = 3cosx - 3sinx = 0 --> to find critical values
3cosx - 3sinx = 0
3cosx = 3sinx
tanx = 1
x = pi/4 and 5pi/4
This splits the domain into [0 , pi/4) ,(pi/4 , 5pi/4) and (5pi/4 , 2pi]
Region 1 : [0 , pi/4)
Testvalue = pi/6
f'(x) = 3cosx - 3sinx
f'(pi/6) = 3cos(pi/6) - 3sin(pi/6)
f'(pi/6) = positive value
So, INCREASING
Region 2 : (pi/4 , 5pi/4)
Testvalue = pi
f'(x) = 3cosx - 3sinx
f'(pi) = 3cos(pi) - 3sin(pi) = -3 --> negative value
So, DECREASING
Region 3 : (5pi/4 , 2pi]
Testvalue = 3pi/2
f'(x) = 3cosx - 3sinx
f'(3pi/2) = 3cos(3pi/2) - 3sin(3pi/2) = 3 --> positive
So, INCREASING
At x = pi/4, increasing changes to decreasing
So, x = pi/4 is a maximum
f(pi/4) = 3sin(pi/4) + 3cos(pi/4)
f(pi/4) = 3/sqrt2 + 3/sqrt2
f(pi/4) = 6/sqrt2
f(pi/4) = 3sqrt(2)
At x = 5pi/4 , decreasing changes to increasing
So, x = 5pi/4 is a minimum
f(5pi/4) = 3sin(5pi/4) + 3cos(5pi/4)
f(5pi/4) = -3/sqrt2 - 3/sqrt2
f(5pi/4) = -6/sqrt2
f(5pi/4) = -3sqrt(2)
So, local maximum value = 3*sqrt(2)
Local minimum value = -3*sqrt(2)
Concavity :
f'(x) = 3cosx - 3sinx
Deriving again :
f''(x) = -3sinx - 3cosx = 0
tanx = -1
x = 3pi/4 and 7pi/4
This splits the domain into [0 , 3pi/4) , (3pi/4 , 7pi/4) and (7pi/4 , 2pi]
Region 1 : [0 , 3pi/4)
Testvalue = pi/2
f''(x) = -3sinx - 3cosx
f''(pi/2) = -3sinpi/2 - 3cospi/2
f''(pi/2) = -3 --> negative
So, CONCAVE DOWN
Region 2 : (3pi/4 , 7pi/4)
Teatvalue = pi
f''(pi) = -3sin(pi) - 3cos(pi)
f''(pi) = 3 --> positive
So, CONCAVE UP
Region 3 : (7pi/4 , 2pi]
Testvalue = 11pi/6
f''(11pi/6) = -3sin(11pi/6) - 3cos(11pi/6)
f''(11pi/6) = 3/2 - 3sqrt3/2 --> negative
So, CONCAVE DOWN
So, here are all of the answers :
Increasing : [0 , pi/4) U (5pi/4 , 2pi]
Decreasing : (pi/4 , 5pi/4)
Local minimum value = -3sqrt(2)
Local maximum value= 3sqrt(2)
Concave up : (3pi/4 , 7pi/4)
Concave down : [0 , 3pi/4) U (7pi/4 , 2pi]