Question

I have a critcal value, which i dtermined is a max. Now i want to plug it back into the original function inorder to get the y value. The orginal function is (x/2)+cosx. The critical point is pi/4.The books gives me an answer of root 2. How did they get this?

Answer 1

hrmm. let me do the calc on it.

Answer 2

derivitive of (x/2) is 1/2 and cosx is -sinx
so 1/2-sinx=0
sinx = 1/2 at pi/6 ? been awhile am i doing that right?
take derivitive set to 0 to find critical points?

Answer 3

if so then pi/6 is max, pi/4 was slightly off of zero slope

Answer 4

still dont get root 2 though

Answer 5

Maybe the book is wrong...?

Answer 6

yeah, that the problem, i cant do the computation, i dont know how to deal with this pi/4 term

Answer 7

are you looking for a global max?

Answer 8

Sin pi/4 = 1/root 2

Answer 9

no , i know that the critical point gives me a rel. max. and now i want to plug the critical point bakc into the orginal equation inorder to get the y value

Answer 10

i get pi/6 if i just solve 1/2=sinx so critical point is pi/6

Answer 11

and y = 1.1278

Answer 12

something like this: \[((\pi/6)/2)+\cos(\pi/6)\]

No, your right the critical point is pi/6, thats what i meant to write, but now i want to find the y value, and to to that i need to plug pi/6 into the orginal equation, which gives me the equation above

Answer 13

The books says the answer should be (pi+6root3)/12)

Answer 14

your book must has to be wrong i guess?

Answer 15

it's asking for the point of the relative max?

Answer 16

I got it now, thanks

Answer 17

oh what was the problem?

Answer 18

laziness

Answer 19

lol... k