What are the absolute extrema of #f(x)=sin2x + cos2x in[0,pi/4]#?

2 Answers
Feb 18, 2017

Answer:

Absolute max: #x = pi/8#
Absolute min. is at the endpoints: #x = 0, x = pi/4#

Explanation:

Find the first derivative using the chain rule:
Let #u = 2x; u' = 2#, so #y = sinu + cos u#

#y' = (cosu) u' - (sinu) u' = 2cos2x - 2sin2x#

Find critical numbers by setting #y' = 0# and factor:
#2(cos2x-sin2x) = 0#

When does #cosu = sinu#? when #u = 45^@ = pi/4#
so #x = u/2 = pi/8#

Find the 2nd derivative: #y''= -4sin2x-4cos2x#

Check to see if you have a max at #pi/8# using the 2nd derivative test:

#y''(pi/8) ~~-5.66 < 0#, therefore #pi/8# is the absolute max in the interval.

Check the endpoints:
#y(0) = 1; y(pi/4) = 1# minimum values

From the graph:
graph{sin(2x) + cos(2x) [-.1, .78539816, -.5, 1.54]}

Feb 18, 2017

Answer:

#0 and sqrt2#. See the illustrative Socratic graph.

Explanation:

graph{(|sin(2x)+cos(2x)|-y)y(y-sqrt2)=0 [-4, 4, 2, 2]}

Use #| sin (theta) | in [0, 1]#.

#|f|=|sin2x+cos2x|#

#sqrt2|sin2x cos(pi/4)+cosx sin(pi/4)|#

#=sqrt2|sin(2x+pi/4)| in [0, sqrt 2]#.