Question

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

Answer 1

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

Answer 2

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

Answer 3

ok

Answer 4

now what?

Answer 5

now you test the critical points and endpoints

Answer 6

i plug them back into the original equation?

Answer 7

the critical point is not 3pi/4

Answer 8

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 #######

Answer 9

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

Answer 10

ok so far? did that make sense

Answer 11

yes

Answer 12

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.

Answer 13

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

Answer 14

and is the lowest of those the minimum?

Answer 15

correct

Answer 16

thanks!