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#