how do i find all the minimum values of x+sin^2 x with the domain of pi/6 to 5pi/6?

Question

perl · Answer

first find the critical points
solve  f ' (x) = 0

anonymous · Answer

i found that the derivative is 1+ 2sincos and that the critical point is 3pi/4

perl · Answer

ok

anonymous · Answer

now what?

perl · Answer

now you test the critical points and endpoints

anonymous · Answer

i plug them back into the original equation?

perl · Answer

the critical point is not 3pi/4

perl · Answer

1 + 2sinx cos x = 0

1 + sin(2x) = 0

sin(2x) = -1

2x= arcsin ( -1)

x =  arcsin(-1) / 2   #######don't forget to divide by 2 #######

perl · Answer

x = ( 3pi/4 + 2pi*n ) / 2 
x = 3pi/8  + pi*n  ,  and since we want domain [pi/6, 5pi/6]  

x = 3pi/8  is our critical point

perl · Answer

ok so far? did that make sense

anonymous · Answer

yes

perl · Answer

ok now we use the Absolute maxima theorem

If f(c) is an absolute maximum/minimum, then c is either a critical point or an endpoint.

perl · Answer

so evaluate
f(pi/6)=
f(3pi/8)=
f(5pi/6) =

anonymous · Answer

and is the lowest of those the minimum?

perl · Answer

correct

anonymous · Answer

thanks!