Question

Find the Min and Max Values

Answer 1

\[2\cos(t) +\sin(2t) [0,\pi/2]\]

Answer 2

I found the derivative, -2sin(t) + 2cos(2t)

Answer 3

buut now im stuck

Answer 4

equate the derivative to 0 and solve for t

Answer 5

right, I guess this is where my trig game is just really weak. could you possibly walk me through that @cwrw238 I'd really appreciate it.

Answer 6

use the identity cos2t = 1 - 2sin^2 t

Answer 7

oh ok! and then solve like a quadratic?

Answer 8

so

-2 sint + 2(1 - 2 sin^t) = 0

yes solve this quadratic

Answer 9

pull a 2 and get (2sint+1) (sint-1) and then solve them separately for zero?

Answer 10

yup - looks right

Answer 11

pi/6 and pi/2 i believe.

Answer 12

i get (2 sint - 1)(sint + 1) = 0

Answer 13

when factoring wouldnt you need a negative sint in the middle, so 2 times the negative?

Answer 14

no i think its + in the middle - ill check again

Answer 15

actually I think you're right..

Answer 16

taking out the 2:

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

Answer 17

yes, that is correct. thank you. so then one of the answers, 3pi/2 wouldnt be in the domain of the function and would therefor be negligible?

Answer 18

sint = 1/2, -1)

Answer 19

right

Answer 20

awesome! ok thank you very much! that was great

Answer 21

yw