Question

Determine whether Rolle's Theorem can be applied to the function on the given interval; if so, find the value(s) of c guaranteed by the theorem. (Enter your answers as a comma-separated list. If Rolle's Theorem does not apply, enter DNE) f(x) = ln(3 − sin(x)) on [0, 2π]

Answer 1

What does the Rolle's Theorem state?

Answer 2

Rolle's Theorem:
1. f(x) is continuous on the closed interval at [a,b]
2. f(x) is differentiable on the open interval (a,b)
3. f(a)=f(b)

If all 3 rules applies then you there should be at least one number,c, such that f'(c)=0

Answer 3

Its like the Mean Value Theorem, but f'(c) should equal to zero instead

Answer 4

Okay. @cpagador can you show that \(\log(3-\sin x)\) is continuous on \([0,2\pi]\)? differentiable? Does \(\log(3-\sin0)=\log(3-\sin2\pi)\)?

Answer 5

Great, now look at the function f(x) first, you should be able to argue f(x) is defined for all x in the interval and check if condition 3 is fulfilled. Then since the ln function is differentiable on its domain, so is f(x). Finally, differentiability implies continuity. :)

Answer 6

@oldrin.bataku yeah, log(3-sin0)=log(3-sin2pi)

Answer 7

@drawar so, what would you do? since log(3-sin0)=log(3-sin2pi), there should be a number where f'(c) would equal to zero?

Answer 8

i had a problem earlier exactly like this but the only difference is, instead of the 3, it was 1. i put that it was DNE and it was correct. i thought that this would be DNE as well but its not.

Answer 9

because f(x)=ln(1-sin(x)) is undefined at 1 while g(x)=ln(3-sin(x)) is not

Answer 10

* at sin(x)=1

Answer 11

uhh... alright then.
sorry.. im still lost on how to solve it. it is a continuous and it i a differentiable like how you said @drawar log(3-sin0)=log(3-sin2pi). the only part im missing is the last part where there should be a number, c, where f'(c)=0.

Answer 12

You now need to find f'(x) of f(x) = ln(3 − sin(x))

Then set f'(c) = 0 and solve for c.

Answer 13

Rolle's Theorem only claims that such a c does exist. If you want to find it explicitly, do as what @agent0smith has said.

Answer 14

Use the chain rule on f(x) = ln(3 − sin(x)) - so differentiate the inside function (the 3-sinx) and multiply it by it by the derivative of ln.

if F(x) = f(g(x))
then F'(x) = f '(g(x)) g '(x)

so g(x) = 3-sinx and f(g(x)) = ln(3-sinx)

Answer 15

@agent0smith alright.. that gonna be 1/(sin(x)-3. correct? D:

Answer 16

You forgot to differentiate the inside function \[\large f(x) = \ln(3 − \sin(x)) \]
derivative of the inside function (3-sinx) is -cosx
\[\large f'(x) = \frac{ 1 }{ 3-\sin x } (-\cos x)\]

Answer 17

oh kilt! yeah, i forgot the -cos(x). -.-"

Answer 18

Now, find c from

\[\Large f'(c) = \frac{ -\cos c }{ 3-\sin c } = 0\]

Answer 19

Solve for c\[\Large \frac{ -\cos c }{ 3-\sin c } = 0 \]
Don't forget the interval [0, 2π]

Answer 20

oh crap, i didnt even see you replied. lol

Answer 21

uhh... i have no clue how to get c. =/

Answer 22

It might look complicated, but it's easy :) What happens if you multiply both sides by (3-sinc)?

\[\Large \frac{ -\cos c }{ (3-\sin c) } = 0\]

Answer 23

it will be -cosc*(3-sinc)=0..? o.o"

Answer 24

Not quite... \[\Large \frac{ -\cos c }{ (3-\sin c) } \times (3 - \sin c) = 0 \times (3-\sin c)\]

Answer 25

thats not the same?wait.. doesn't the (3-sinc) on the left side cancel out? and on the right it will still equal to 0?

Answer 26

Correct... so now you just have...

\[\large -\cos c =0\]

Answer 27

hmm... alright then.

Answer 28

so then if you're solving for c, don't you divide the -cos so that you can
isolate c?

Answer 29

Think of it this way
\[\large \frac{ a }{ b } = 0\]
The only way this can be zero, is if a=0. No value of b can make this equal to zero.

Answer 30

first divide or multiply both sides by -1
\[\large \cos c =0 \]

\[\large c= \cos^{-1} 0\]

Answer 31

that will equal about 1.57

Answer 32

It will, but you can get it from a unit circle or table of values http://www.sciencedigest.org/UnitCircle.gif

Answer 33

so since its on the interval at [0.2pi], then it will be at pi/2.
correct?

Answer 34

Note that there are 2 values of c, you can see both on the unit circle, where cos = 0

Answer 35

ohhhhh! alright. so pi/2 AND at 3pi/2.
so since its from [0,2pi] it has two values? hmm.. alright then!

Answer 36

i should've came here a looong time ago. i missed the submission time of this hw and this was the only problem i was missing from the hw. but, its fine. this helped me so much! i learned more here than in my actual calc class. haha. thanks so much esp @agent0smith . :)

Answer 37

No prob :)

Answer 38

so pi/2 AND at 3pi/2

^ correct.