Question

Absolute minimum and maximum values of
`f(x)=2cos (x) +sin (2x)` on the interval
`[0,pi/2]`

Answer 1

Explanation:

f'(x)=-2sin(x)+2cos(2x)=0

as cos(2x)=2sin(x)cos(x),

-2sin(x)+4cos(x)sin(x)=0

sin(x)-2cos(x)sin(x)=0

(sin(x))(1-2cos(x))=0

-> x = 0, pi/3

testing these values along with the end points of the interval,

f(0)=2

f(pi/3)=1+(0.5sqrt(3))

f(pi/2)=0

so the min is 0 and the max is 2.