Question

How can i find the maxima and minima using differentiation?
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Answer 1

i mean i know i have to find the values of f'x=0, the f''x=0...
f'x=0 gives me cosx=0 and sinx=2... doesnt make sense..

Answer 2

sinx=-2*

Answer 3

@.Sam. This seems to be right up your ally

Answer 4

@sarah_98 :P

Answer 5

how do you expect me to know? xD

Answer 6

show me your derivatives

Answer 7

eh, dont tag gofor. :P

Answer 8

f'(x)= -8cos(x) - 4cos(x)*[-sin(x)]

Answer 9

so yeah i was wrong.. sinx=2 .. but thats not possible still..

Answer 10

f = 11 - 8sin - 2cos^2

f' = - 8cos + 4 sin cos
= - 8cos + 2 sin(2x)

Answer 11

right. next?

Answer 12

i spose we could play with

0 = - 8cos + 4 sin cos ; is already true for cos=0, odd multiples of pi/2

8cos = 4 sin cos

2cos = sin cos

2 = sin ; which is never .... so id say thats a dummy variable

Answer 13

so that means there is either a max or a min.. not both?

Answer 14

might be both, we havent done an f''

is there a specific interval for this? 0 to 2pi perhaps?

Answer 15

umm. i dont know.. thats the question.
i have the solution to it.. they have evaluated both max and min using -1<=sinx<=1.. so must be [0,2pi]..

Answer 16

eh, that doesnt mean [0 2pi] necessarily.. ignore that.

Answer 17

then lets use pi/2 and 3pi/2 as critical points to evaluate

Answer 18

ok so f' = - 8cos + 2 sin(2x)... f''=8six + 4cos(2x)..

Answer 19

f'' is not really needed

Answer 20

umm. why? how?

Answer 21

well, the critical points we have at the moment can be used to see if x=pi/2 is bigger or smaller than x=3pi/2

f'' would simply be extra work that is not required of us