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

1 Answer

5 votes

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.

User AbdulAhmad Matin
by
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