Question

what are the minimum and maximum values of x+sin^2 x with the domain of pi/6 to 5pi/6?

Answer 1

f(x) = x + sin^2(x)

to find max/min, find the critical points, then use second derivative test

Answer 2

is this a closed domain?

Answer 3

its greater than or equal to

Answer 4

pi/6 <= x <= 5pi/6 ?

Answer 5

yes

Answer 6

ok then we can find the absolute max and min (which means we need to test the endpoints as well)

Answer 7

go ahead and find the derivative, and set it equal to zero

Answer 8

i found that the first derivative 1+2sincos

Answer 9

ok, and 2sinx cosx is equal to sin(2x) by a trig identity

Answer 10

f ' (x) = 0

1 + 2 sin x cos x = 0

1 + sin(2x) = 0

sin(2x) = -1

Answer 11

what is sin(2x)=-1?

Answer 12

that is for f' =0 when sin (2x) =-1, solve for x is the next step

Answer 13

we can take inverse sine

Answer 14

which is cos

Answer 15

not quite. i meant like arcsine

Answer 16

sin(2x)=-1
<=>
2x = sin^-1 ( 2x )

Answer 17

so, when sin (something ) =-1? what is that "something"?

Answer 18

3pi/2, right? that is sin (3pi/2) =-1, right?

Answer 19

right

Answer 20

now, our "something" is 2x , so that 2x = 3pi/2, right? then x = ???

Answer 21

3pi?

Answer 22

hey, 2x = 3pi/2, x = 3pi/2/2= 3pi/4 hahaha.... take a rest, friend.

Answer 23

no??