Find the relative extrema

Question

Find the relative extrema for f(x)=2sin^2x + cos2x on the interval [0,2π]. Also find the x coordinates of the inflection points of f(x) on this interval. Give the intervals where the function is increasing, where it is decreasing...Sketch the function.

Answer 1

f(x) = 2sin^2(x) + cos2x from x=-0 to 2pi

find extrema and inflection points, intervals increasing and decreasing

or did you mean f(x) = 2sin^2(x) = cos^2(x)

following takes the 1st interpretation

take the derivative and set = 0

f'(x) = 4cosx - 2sin 2x = 0

2cosx -sin2x = 0

2cosx = 2sinxcosx

cosx = sinxcosx

cosx = 0 sinx = 1

x = pi/2, 3pi/2 are where f(x) is a global max (2.09, 2)

f(pi/2) = 2

f(3pi/2) = 2

x = about 2.63 radians when f(x) is at a global minimum = (2.63, 0)

also relative max = (pi/12, 1.63)

and relative min = (5pi/12, .37)

increasing interval (0, pi/12), (5pi/12, 2.09) (2.63, 2pi)

decreasing intervals (pi/12, 5pi/12), (2.09, 2.63)

four inflection points, about midway between max and min points

average the x coordinates and the y coordinates of each of the 4 extrema

about (3pi/12, 1), ((5pi/12 +2.63)/2, (2.09+2pi)/2),

use a graphing calculator such as Desmos online to graph the function

using the 2nd intepretation of the problem as

f(x) = 2sin^2(x) + cos^2(x)

then

= sin^2(x) + sin^2(x) + cos^2(x)

= sin^2(x) + 1 which just vertically shifts the sine curve up by 1

global & relative minimums are (0,1), (pi,1) (2pi, 1)

global and relative maximums are (pi/2, 2) (3pi/2, 2)

inflection points are (pi/4, 3/2) (3pi/4, 3/2) (5pi/4, 3/2) (7pi/4, 3/2)

increasing intervals (0,pi/2) (pi, 3pi/2)

decreasing intervals (pi/2, pi), (3pi/2, 2pi)

graph looks just like a basic sine curve but shifted up by 1 unit