How Does the Rolle Theorem Apply to Continuous and Differentiable Functions?

[haya]

Homework Statement

the rolle theorem said :
suppose
(i) F(X) is continuous on colse interval (a,b)

(ii) F(X) is differentiable on open interval ( a,b)

(iii) F(a)=F(b)

then there is c on (a,b) such that F`(c)=0


the question is give an example :

1- satisfied (i) (ii) and not satisfy (i) and explain why c is not on the interval (a,b)?

2-satisfied (i) (iii) and not satisfy (ii) and explain why c is not on the interval (a,b)?

3-satisfied (ii) (iii) and not satisfy (i) and explain why c is not on the interval (a,b)?


I got answers each 1 and 2 but on 3 how can I got it
because if the function discontinous then so it is non diffrentiable on (a,b).

 

[HallsofIvy]

Homework Statement

the rolle theorem said :
suppose
(i) F(X) is continuous on colse interval (a,b)

That's not a closed interval. You mean [a, b].

(ii) F(X) is differentiable on open interval ( a,b)

(iii) F(a)=F(b)

then there is c on (a,b) such that F`(c)=0


the question is give an example :

1- satisfied (i) (ii) and not satisfy (i) and explain why c is not on the interval (a,b)?

2-satisfied (i) (iii) and not satisfy (ii) and explain why c is not on the interval (a,b)?

3-satisfied (ii) (iii) and not satisfy (i) and explain why c is not on the interval (a,b)?


I got answers each 1 and 2 but on 3 how can I got it
because if the function discontinous then so it is non diffrentiable on (a,b).

But, as I pointed out above that is NOT a closed interval. Can you think of a function that is differentiable on (a, b) but NOT continuous on [a, b]? Of course, that depends precisely upon the difference between (a, b) and [a, b].

 

[haya]

yes I mean closed interval [a,b]


for example F(x)= 1/x on [0,2]

F(x) is discontinuos at x=0

F(x) is differentiable on (0,2)


and thank you so much

 

[Pere Callahan]

for example F(x)= 1/x on [0,2]

This example does not satisfy (iii) since $?=F(0)\neq F(2)=1/2$

 

[HallsofIvy]

Try F(x)= 1/x on (0, 2], F(0)= 1/2.

 

[haya]

yes it is satisfyed the theorem

 

[haya]

I still thinking about this question


but we know if the function is not continuous then it is not diffrentiable


so the answer of question (3) is will be no example because he want the function be discontinuous on interval [a,b] and diffrentiable on (a,b) that's never ever will happen in mathematics

 

[Pere Callahan]

The function in our example is continuous everywhere except at zero. At zero however, it is not required to be differentiable, so no problem.

 

[haya]

could you give other example :

the example will be in discontinuoty on [a,b]
but differentiable on (a,b)

ok

I will be greatfull for you if give me the example
because this qeustion from the homework and tomorrow is the due day

 

[Office_Shredder]

You're correct, to the point where if the function is differentiable on (a,b) it's continuous on (a,b). Hence the only discontinuities can be at the endpoint. Examples are usually just continuous functions on (a,b) with the endpoints f(a) and f(b) redefined. Such as (a=0,b=1)

f(x)=x on (0,1) f(0)=f(1)=1
f(x)=sin(x) on (0,1) f(0)=2 f(1)=-3
f(x)=e^x on (0,1) f(0)=1 f(1)=0