The minimum is #g(pi-sin^(-1)(2/5)) ~~0.8776#
The maximum is #g(2pi) ~~ 17.5663#
To find absolute extrema of a function that is continuous on a closed interval: find critical numbers that are in the interval, evaluate the function at the endpoints and at the critical numbers
You'll want some technology to finish the arithmetic.
#g(x) = 2x + 5cosx# on the interval #[0,2pi]#
#g'(x) = 2 - 5sinx = 0# where #sinx = 2/5#
I do not recognize a value of #x# that will work, but I do know that there are 2 such value in the interval #[0, 2 pi]#. One is in the first quadrant (call it #x_1#) and one is in the second (#x_2#).
Using #sin^2x+cos^2x=1# we can work out that at #x_1# we also have
#cosx_1 = sqrt(1-sin^2x_1) = sqrt(1-(2/5)^2) = sqrt (21/25) = sqrt21/5#.
At the other, #cosx_2 = - sqrt21/5#.
Now, we need to find #g(x)# for four numbers:
#g(0) = 2(0) + 5cos(0) = 5#
#g(x_1) = 2(x_1) + 5cos(x_1) = 2(x_1) + 5(sqrt21/5)=2(x_1) + sqrt21#
#g(x_2) = 2(x_2) + 5cos(x_2) = 2(x_2) + 5( - sqrt21/5)=2(x_2) - sqrt21#
#g(2pi) = 2(2pi) + 5cos(2pi) = 4pi + 5#
Using technology to find the values of #x_1# and #x_2#
With #x_1 ~~ 0.4115# (so #x_2 = pi - x_1 ~~ 2.7301#)
and #sqrt21 ~~ 4.5826#,
We get
#g(0) == 5#
#g(x_1) ~~ 5.4056#
#g(x_2) ~~0.8776#
#g(2pi) ~~ 17.5663#
The minimum is #g(x_2) ~~0.8776#
The maximum is #g(2pi) ~~ 17.5663#
