Question

Show whether Rolle's Theorem can be applied to f(x)

If it can, determine any values of c in the interval [0,2pi] for which f'(c)=0

f(x)=cos x

Answer 1

wats the hypothesis of Rolle's theorem ?

Answer 2

we need to show/state f(x) = cosx satisfies the hypothesis of Rolle's theorem right ?

Answer 3

yes, to show it can be applied. hold on. I'll post it up

Answer 4

take ur time :)

Answer 5

Let f be continuous on the closed interval [a,b] and be differentiable on the open interval (a,b). If: f(a)=f(b), then there is at least one number c in (a,b) such that f'(c)=0

Answer 6

thats it :)

Answer 7

Yes, lets see if f(x) satisfies the conditions required for applying Rolle's theorem :-

```
Let f be continuous on the closed interval [a,b] and be differentiable on the open interval (a,b). If: f(a)=f(b),
```

Answer 8

\(f(x) =\cos x\)

\(f(x)\) is differentiable in \((0, 2\pi)\)

Also,

\(f(0) = f(2\pi) = 1\)

So, the hypothesis of Rollle's theorem are satisfied.

Answer 9

next, find f'(x) and set it equal to 0 and solve

Answer 10

f'(x) = -sin(x) = 0

x = ?

Answer 11

positive sin?

Answer 12

x=sin ?

Answer 13

f'(x) = -sin(x) = 0

x = arcsin(0)
= 0, pi, 2pi

Answer 14

so the values of c for which f'(c) = 0 in interval [0,2pi] are : c = 0, pi, 2pi

Answer 15

oh right. forgot thats how you did it

Answer 16

:)

Answer 17

ganeshie you are the best

Answer 18

thank you so much

Answer 19

np :)

Answer 20

just a quick question :-
Is the given interval in question really closed ? [0,2pi]

Answer 21

usually, in questions involving Rolle's theorem, they give open interval (0, 2pi)

Answer 22

yeah it is. why?

Answer 23

no it has the brackets

Answer 24

if it is open interval the answer changes, thats why asking...

Answer 25

[ ] those little guys

Answer 26

ah I see

Answer 27

if it is open interval, answer wud be just : c = pi
okay cool :) just want u see thats all

Answer 28

got it. Thanks. I have another question can you help me with it?

Answer 29

sure... il try post it :)