Consider the equation below. (If an answer does not exist, enter DNE.) f ( x ) =
ID: 2851401 • Letter: C
Question
Consider the equation below. (If an answer does not exist, enter DNE.)
f(x) = 2cos2(x)4sin(x), 0x2
(a) 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.)
(_______)
(b) Find the local minimum and maximum values of f.
(c) Find the inflection points.
(x, y) = (_______)
(smaller x-value)
(x, y) = (_______)
(larger x-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.)
(_______)
local minimum value (_______) local maximum value (_______)Explanation / Answer
f(x) = 2cos2(x)4sin(x)
==> f '(x) = 2(2cosx)(-sinx) - 4cosx
==> f '(x) = -4cosxsinx - 4cosx
critical points ==> f '(x) = 0
==> -4cosxsinx - 4cosx = 0
==> -4cosx(sinx +1) = 0
==> cosx = 0 , sinx = -1
==> x = /2 , 3/2
function incresing ==> f '(x) > 0
==> x belongs to (/2 , 3/2 )
function decreasing ==> f '(x) < 0
==> x belongs to (0 , /2) U (3/2, 2)
f(/2) = 2cos2(/2)4sin(/2) = -4
f(3/2) = 2cos2(3/2)4sin(3/2) = 4
Hence local minimum value is -4
Hence local maximum value is 4
f '(x) = -4cosxsinx - 4cosx = -2sin(2x) -4cosx
==> f ''(x) = -2(cos(2x) (2)) -4(-sinx) = -4cos(2x) +4sinx
inflection points ==> f ''(x) = 0
==> -4cos(2x) +4sinx = 0
==> -4(1 -2sin2x) + 4sinx = 0
==> -4 +8sin2x + 4sinx = 0
==> 2sin2x + sinx -1 = 0
==> sinx = [-1 + (1 -4(2)(-1))]/2(2) , [-1 - (1 -4(2)(-1))]/2(2)
==> sinx = [-1 + 3]/2(2) , [-1 - 3]/2(2)
==> sinx = 1/2 , -1
==> x = /6 , 5/6
Hence concave up ==> f ''(x) > 0
==> x belongs to (/6 , 5/6 )
concave down ==> f ''(x) < 0
==> x belongs to (0, /6) U (5/6 ,2)