KaySolve Rolle's Theorem Homework: Check Validity & Continuity

[macjack]

Homework Statement

check whether the rolle's theorem is valid
f(x)= x^2 in [1,2]
f(x)= 1-(x-1)^(2/3) in [0,2]

my question is, how do you usually find whether these equations are differentiable/continuous ??

For some equation, we used to plot the graph and find any corner points are there or not and from there we can say whether it is differentiable and inturn it is continuous or not ??

when i read that in the book , it just said it is differentiable and continuous but i didnt say how they come to that conculsion ! can you please help me out to find that ??

I know the relation between the differentiation and continuous !

Thanks
Mc

 

[Dick]

If by Rolles theorem you mean f(a)=f(b) implies f'(x)=0 for some x in [a,b], subject to conditions on f, I would ask you for the first f, is f(a)=f(b)? And for the second f, is f differentiable at x=1? Take the derivative and study values of x near 1.

 

[macjack]

Thanks for the reply. For the above two equations i could able to prove that it is not satisfying rolle's theorem.For the first one f(a) Not equal to f(b) and for the second one f'(x) is infinity.

BTW.. if i change the range from [1,2] to [-1,1](for the first f), f(a)= f(b) then do i have to differentiate and then apply the values to see whether it is differentiable or not ?

what i wanted to know is , is there any common procedure/tip do i have to understand to disprove rolle's theorem for some f ?

say for f(x)= |x| in [-1,1], f(a)=f(b) , then we plotted some graph to find whether it is differentiable or not right ??

Am i missing something ?

Thanks again for your reply

 

[lalbatros]

Rolles theorem is always valid since it has been proven.
But Rolles theorem may not be applicable if the function does not satisfy Rolles theorem conditions of applicability.

 

[Office_Shredder]

mackjack, the proper method of showing whether a function is differentiable or not is showing $$lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$$ exists for all x in the interval. Usually, this is pretty simple. In cases like |x|, it helps to know there's a corner at 0, because you can show that the limit from the left hand side doesn't equal the limit from the right hand side, so the limit just doesn't exist

 

[Dick]

I think in this case there isn't a need for a formal 'proof' of differentiability. E.g. taking x^2 on the interval [-1,1], the derivative is 2*x which is a well behaved function on [-1,1] (and has a zero as rolle's theorem demands). What about your second f?

 

[macjack]

thanks

Office_Shredder, thanks for the explanation. Yes i found out the limits (both left and right)for my latest problem |x| in range [-1,1]. It's not same.

Now let me come to the problem, you found that |x| is not differentiable at x=0.How did you find that ? (We have seen this example in sooo many places)

If somebody gives me a f(x) that is a polynomial function/or some function and a range [a,b] and i found that f(a)=f(b). Now i have to find whether it is differentiable or not !

What do you do in this case ?
1. Differentiate and then apply some value in that range for x.
2. Draw a graph and see is there any corner points ?
3. ...?

For example,when i gave this problem,
f(x)= 1-(x-1)^(2/3) in [0,2]

Dick gave me a hint saying it is not differentiable at x=1 ! Now i want to know ,how did he come to the conculsion that, it is not differtiable at x=1 just by seeing the problem ?

Am i confusing ?

Thanks
Mc

 

[Dick]

What is the expression for the derivative of 1-(x-1)^(2/3)? What happens if x=1? What happens if x is near 1?

 

[macjack]

am i correct ?

f(x) = 1- (x-1)^(2/3)

so f'(x) = 0 - (2/3) (x-1)^(-1/3)
= (-2/3) / ((x-1)^(1/3))

so if x=1, bottom x-1 will become 0 and result becomes infinity.

if x<1, then you will get a positive value and if you x>1 you will get a negative value !

So this function is not satisfying the three conditions of rolles theorem.
deck, how did you find that it is discontinous at x=1 ? is it because you see a (x-1) in the function f ?

 

[Dick]

f(x) = 1- (x-1)^(2/3)

so f'(x) = 0 - (2/3) (x-1)^(-1/3)
= (-2/3) / ((x-1)^(1/3))

so if x=1, bottom x-1 will become 0 and result becomes infinity.

if x<1, then you will get a positive value and if you x>1 you will get a negative value !

So this function is not satisfying the three conditions of rolles theorem.
deck, how did you find that it is discontinous at x=1 ? is it because you see a (x-1) in the function f ?

It IS continuous at x=1 (f(1)=0), it's not differentiable at x=1. To find this out I did the same thing you just did, I looked at the derivative and saw an x-1 in the denominator. It's a 'cusp'.