Question

Mean Value theorem question...(posting below)

Answer 1

a = 0, b = pi

Answer 2

So find f(0) and f(pi)

Answer 3

I know that it is continuous, and I know that the outputs for both the intervals are 0. So (0,0) and (pi,0).

Answer 4

Would I just find the the slope with those coordinates now?

Answer 5

That formula is basically the slope formula, so what is f(0)?

Answer 6

It is 0

Answer 7

Right and f(pi)?

Answer 8

0

Answer 9

So your answer is?

Answer 10

0 - 0
-----
pi - 0

so 0/pi or just 0

Answer 11

Right, but now, what is the derivative of that function?

Answer 12

cos(x) + 2cos(2x)

Answer 13

Good, now equate it with 0

Answer 14

Since 0 is what you got from your formula

Answer 15

cos(x) + 2cos(2x) = 0
2cos(2x) = -cos(x)
(1/2)(sec(2x)) = -sec(x)

I'm not sure what to do next

Answer 16

Hey sorry back

Answer 17

Ok so, the derivative is 2cosx+2cos(2x) since 2 is constant

Answer 18

Now we have \[cosx+\cos(2x) = 0 \] use the identity \[\cos(2x) = 2\cos^2(x)-1\] that should make it simpler for you

Answer 19

You should notice something once you use that identity

Answer 20

Sorry, I walked away for a minute. I'm back now.

Answer 21

I'm confused on how you got rid of the 2 (the constant).

Answer 22

divide both sides by 2

Answer 23

Oh yay. I misread.

Answer 24

there is one mistake in the derivative of f

Answer 25

What's that?

Answer 26

oh @Astrophysics found it

Answer 27

Haha

Answer 28

Thanks for checking freckles!

Answer 29

Ok so once you use the identity we get \[cosx+2\cos^2x-1=0\]

Answer 30

Reminds you of anything?

Answer 31

Um, I'm sorry, but I'm lost. D:
Just a second. I'll look back over it and see if I missed something.

Answer 32

Okay, just checking...where does the -1 come from?

Answer 33

Hint: \[ax^2+bx+c=0\]

Answer 34

Remember I used the identity \[\cos(2x) = \color{red}{2\cos^2x-1}\]

Answer 35

I'm so sorry, but I don't know what I should do. If these were numbers, I would understand, but I'm terrible with trigs

Answer 36

It's alright, so you understand how I got to this part \[cosx+2\cos^2x-1=0\] right? Let me put it in a better way, \[2\cos^2x+cosx-1=0\]

Answer 37

Now you can factor

Answer 38

\[\left(\cos \left(x\right)+1\right)\left(2\cos \left(x\right)-1\right)\]

Answer 39

yes exactly

Answer 40

Now solve for x

Answer 41

Should I set both factor pieces equal to 0?

Answer 42

yes

Answer 43

cos x = -1
cos x = 1/2

Answer 44

Looks good

Answer 45

Keep going

Answer 46

How would I get the cos away from the x?

Answer 47

You can take the inverse but we have c = pi, and pi/3 right?

Answer 48

Yes

Answer 49

So I should just replace the cos with their c values?

Answer 50

the x

Answer 51

cos (pi/3) = 1/2
cos (pi) = -1

Answer 52

1/2 = 1/2 = 0
-1 = -1 = 0

Answer 53

Those are you c values in the interval, c = pi, and pi/3

Answer 54

Oh so those are the answers?

Answer 55

Is that what you meant?

Answer 56

Correct