f(t)=16cost + 8sin2t [0 , pi/2] I have gotten the minimum value which is 0, howe

Question

f(t)=16cost + 8sin2t      [0 , pi/2]

I have gotten the minimum value which is 0, however I keep getting 16 as my maximum value and it is incorrect.

Here is my work:

16cost + 8sin2t ->f '(t) -> chain rule -> 8(2cost + 2(sin2t) -> double angle -> 8(2cost + 2(2sintcost))=0 -> distribute -> 8(2cost + 4sintcost) =0 -> solve for t ->

0/8 is undefined

cost(2+4sint) =0 cost=0 --> pi/2 and 3pi/2

2+4sint=0 -> sint= -1/2 --> 5pi/6 and 11pi/6

All values except for pi/2 are not included in domain so then I plugged that as well as my endpoints into f(t) and got 0 and 16.

I am wondering where I went wrong.

Explanation / Answer

take derivative :

-16sint + 16cos2t = 0

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

sole as quadratic

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