Question

Given h(x)=2cos x-sin 2x, according to the First Derivative Test for what values of x does h have a relative minimum on the interval (-pi,pi)?

x=-5pi/6
x=-5pi/6,-pi/6
x=-5pi/6,-pi/6,pi/2
x=pi/2

What are the points of inflection of k(x)=sin -1/4 sin 2x?
x=-pi/3
x=-pi/6
x=-pi/3,pi/x
x=-pi/3,0,pi/3

Answer 1

For the second one I know it is either C or D.

Answer 2

@Loser66
@welshfella

Answer 3

First find the derivative of cos x - sin 2x
can you do that?

Answer 4

sorry i have to go right now but I see Loser66 is here I'm sure he'll be able to help

Answer 5

The first derivative of h(x) is ?? your answer above is not correct

Answer 6

he gave me a different function of h(x) he dropped the 2 from the front f the cos

Answer 7

type it here, please

Answer 8

the derivative of h(x) is -2cos 2x - 2 sin x

Answer 9

I meant the original h(x) =?

Answer 10

2 cos x-sin 2x

Answer 11

Am I correct for the second question with c or d?

Answer 12

solve one by one, please, the first one, h(x) = 2 cos x - sin(2x), sure?

Answer 13

yeah

Answer 14

h'(x) = (2cos(x) )' - (sin(2x))' = -2sin(x) - 2cos (2x)
Now, set it =0 and solve for x. What do you get?

Answer 15

I dont understand?

Answer 16

When I plug it into a calculator it says ERROR

Answer 17

no way to solve by calculator, you must solve it by hand

Answer 18

can you walk me through the steps?

Answer 19

I'm sorry I never had pre calc

Answer 20

What?? no precal? how to solve cal problem?

Answer 21

They made me skip it guidance counselor told me a could handle it and i am barely passing

Answer 22

ok

Answer 23

ok so can you walk me through it ?

Answer 24

I don't, but wolfram can help. read it carefully. http://www.wolframalpha.com/input/?i=-2sinx+-+2cos%282x%29+%3D0+

Answer 25

So which is the answer? I am leaning towards C

Answer 26

last one, x =pi/2

Answer 27

ok thank you now to the second one. I think it is c or d I cant figure out whether 0 is included or not

Answer 28

@jim_thompson5910 @pooja195 @TheSmartOne

Answer 29

The second one has \[\Large k(x) = \sin(x) - \frac{1}{4}\sin(2x)\] right? or no?