Question

Find the absolute maximum and absolute minimum values of f on the given interval.
f(t) = 2 cos t + sin 2t, [0, π/2]

Answer 1

@ganeshie8

Answer 2

-2sint+2cos2t is the derivativ>

Answer 3

yes, next what

Answer 4

then i convern the 2(cos2t) into sin....so would that be 1-sin^2(sx)

Answer 5

sin^2(2x) i mean

Answer 6

set the first derivative equal to 0 and solve \(t\) in the given interval

Answer 7

but how

Answer 8

\(\large -2\sin t+2\cos(2t) = 0\)

Answer 9

we use this trig identity :
\(\large \cos (2x) = 1-2\sin^2x\)

Answer 10

-sint+cos(2t)=0

Answer 11

oh ok

Answer 12

\(\large -2\sin t+2\cos(2t) = 0\)

\(\large -\sin t+\cos(2t) = 0\)

\(\large -\sin t+1-2\sin^2t = 0\)

\(\large 2\sin^2t + \sin t -1= 0\)

Answer 13

see if u can factor it

Answer 14

sin t = -1 sin t =1 ?

Answer 15

so - pi/2 and positive pi/2

but only positive because its in range?

Answer 16

\(\large 2\sin^2t + \sin t -1= 0\)

\(\large 2x^2 + x - 1= 0\)

\(\large 2x^2 + 2x-x - 1= 0\)

\(\large 2x(x + 1)-1(x + 1)= 0\)

\(\large (x+1)(2x-1)= 0\)

x = -1, 1/2

\(\large \sin t = -1\) or \(\large \sin t = 1/2\)

Answer 17

\(\large t = \pi/6\)

Answer 18

oh ok
i see

Answer 19

So you need to test your function at \(t = \pi/6\) and the boundary values

Answer 20

\(\large f(t) = 2 \cos t + \sin (2t)\)

f(pi/6) = ?
f(0) = ?
f(pi/2) = ?

Answer 21

maximum of above 3 values is the `absolute maximum` in that interval

minimum of above 3 values is the `absolute minimum` in that interval

Answer 22

thank you

Answer 23

yw :)