Post History

50%

+1 −1

Q&A Rules of checking differentiability for Rolle's theorem

0 answers · posted 3y ago by deleted user · edited 3y ago by deleted user

#3: Post edited by (deleted user) · 2021-07-10T06:06:59Z (over 3 years ago)
improve formatting

Copy Link

Raw

Markdown

I was doing some exercises of Rolle's theorem. But, they didn't check the differentiability the way we checked differentiability normally. I am giving some examples.
When I was checking differentiability with limit (before differentiation) I was just putting $x$ values which given in question. Like $x = 2$ check it is differentiable or not.
~~$$Rf'(x)=\lim_{h->0} \frac{f(x-h)-f(x)}{h}$$~~
I had put $2$ instead of $x$ . But, something else is happening in Rolle's theorem.
~~>Verify Rolle's theorem for the function $f (x) = x^{2}$ in the interval [-1,1]~~
In answer they didn't change $x$ .
~~>Verify Rolle's theorem for the function $f(x)=x^{\frac{2}{3}}$ in the interval (-1,1).~~
~~They had written $0$ instead of x this time. I am confused how to find the value. In earlier problem it was closed interval. And, it is open interval now. Which law should I apply to find the value?~~
>For the function $f(x)=x when 0<=x<=1$ $f(x)=2-x when 1<x<=2$ f(0)=0=f(2), according to the Rolle's theorem there exist at least one point in the interval (0,2) such that $f^{'} (x) = 0$ but $f'(x)!=0 $f o r a n y$ x element (0,2)$ why not, explain.
This time they had put $1$ instead of $x$ .
<hr/>
I was solving some problems by myself. I noticed in question they gave us two value inside interval. I was trying to solve first problem (which I wrote here) with $- 1$ instead of $x$ (as I said they used $x$ but, I tried to use $x$ ) than, I noticed that function is differentiable. To verify Rolle's theorem I needed a $x$ for that function. For Rolle's condition $f^{'} (c) = 0$ . Without $x$ I can't solve the condition. Now, I think it's ok if I use $x$ always, isn't it?

I was doing some exercises of Rolle's theorem. But, they didn't check the differentiability the way we checked differentiability normally. I am giving some examples.
When I was checking differentiability with limit (before differentiation) I was just putting $x$ values which given in question. Like $x = 2$ check it is differentiable or not.
$$Rf'(x)=\lim_{h \to 0} \frac{f(x-h)-f(x)}{h}$$
I had put $2$ instead of $x$ . But, something else is happening in Rolle's theorem.
>Verify Rolle's theorem for the function $f(x) = x^2$ in the interval $[-1,1]$.
In answer they didn't change $x$ .
>Verify Rolle's theorem for the function $f(x) = x^{\frac{2}{3}}$ in the interval $(-1,1)$.
They had written $0$ instead of $x$ this time. I am confused how to find the value. In earlier problem it was closed interval. And, it is open interval now. Which law should I apply to find the value?
>For the function $f(x)=x$ when $0 \leq x \leq 1$ ,
$f(x) = 2-x$ when $1 < x \leq 2$ , $f(0)=0=f(2)$, according to the Rolle's theorem there exist at least one point in the interval $(0,2)$ such that $f^{'} (x) = 0$ but $f'(x) \neq 0 $f o r a n y$ x \in (0,2)$ why not, explain.
This time they had put $1$ instead of $x$ .
<hr/>
I was solving some problems by myself. I noticed in question they gave us two value inside interval. I was trying to solve first problem (which I wrote here) with $- 1$ instead of $x$ (as I said they used $x$ but, I tried to use $x$ ) than, I noticed that function is differentiable. To verify Rolle's theorem I needed a $x$ for that function. For Rolle's condition $f^{'} (c) = 0$ . Without $x$ I can't solve the condition. Now, I think it's ok if I use $x$ always, isn't it?

#2: Post edited by (deleted user) · 2021-07-10T06:05:55Z (over 3 years ago)

Copy Link

Raw

Markdown

I was doing some exercises of Rolle's theorem. But, they didn't check the differentiability the way we checked differentiability normally. I am giving some examples.
When I was checking differentiability with limit (before differentiation) I was just putting $x$ values which given in question. Like $x = 2$ check it is differentiable or not.
$R f^{'} (x) = lim_{h - > 0} \frac{f (x - h) - f (x)}{h}$
I had put $2$ instead of $x$ . But, something else is happening in Rolle's theorem.
>Verify Rolle's theorem for the function $f (x) = x^{2}$ in the interval [-1,1]
In answer they didn't change $x$ .
>Verify Rolle's theorem for the function $f (x) = x^{\frac{2}{3}}$ in the interval (-1,1).
They had written $0$ instead of x this time. I am confused how to find the value. In earlier problem it was closed interval. And, it is open interval now. Which law should I apply to find the value?
>For the function $f (x) = x w h e n 0 <= x <= 1$ $f (x) = 2 - x w h e n 1 < x <= 2$ f(0)=0=f(2), according to the Rolle's theorem there exist at least one point in the interval (0,2) such that $f^{'} (x) = 0$ but $f^{'} (x)! = 0$ for any $x e l e m e n t (0, 2)$ why not, explain.
~~This time they had put $1$ instead of $x$ .~~

I was doing some exercises of Rolle's theorem. But, they didn't check the differentiability the way we checked differentiability normally. I am giving some examples.
When I was checking differentiability with limit (before differentiation) I was just putting $x$ values which given in question. Like $x = 2$ check it is differentiable or not.
$R f^{'} (x) = lim_{h - > 0} \frac{f (x - h) - f (x)}{h}$
I had put $2$ instead of $x$ . But, something else is happening in Rolle's theorem.
>Verify Rolle's theorem for the function $f (x) = x^{2}$ in the interval [-1,1]
In answer they didn't change $x$ .
>Verify Rolle's theorem for the function $f (x) = x^{\frac{2}{3}}$ in the interval (-1,1).
They had written $0$ instead of x this time. I am confused how to find the value. In earlier problem it was closed interval. And, it is open interval now. Which law should I apply to find the value?
>For the function $f (x) = x w h e n 0 <= x <= 1$ $f (x) = 2 - x w h e n 1 < x <= 2$ f(0)=0=f(2), according to the Rolle's theorem there exist at least one point in the interval (0,2) such that $f^{'} (x) = 0$ but $f^{'} (x)! = 0$ for any $x e l e m e n t (0, 2)$ why not, explain.
This time they had put $1$ instead of $x$ .
<hr/>
I was solving some problems by myself. I noticed in question they gave us two value inside interval. I was trying to solve first problem (which I wrote here) with $- 1$ instead of $x$ (as I said they used $x$ but, I tried to use $x$ ) than, I noticed that function is differentiable. To verify Rolle's theorem I needed a $x$ for that function. For Rolle's condition $f^{'} (c) = 0$ . Without $x$ I can't solve the condition. Now, I think it's ok if I use $x$ always, isn't it?

#1: Initial revision by (deleted user) · 2021-07-10T05:48:27Z (over 3 years ago)

Copy Link

Raw

Markdown

Rules of checking differentiability for Rolle's theorem

I was doing some exercises of Rolle's theorem. But, they didn't check the differentiability the way we checked differentiability normally. I am giving some examples. 

When I was checking differentiability with limit (before differentiation) I was just putting $x$ values which given in question. Like $x=2$ check it is differentiable or not. 

$$Rf'(x)=\lim_{h->0} \frac{f(x-h)-f(x)}{h}$$

I had put $2$ instead of $x$. But, something else is happening in Rolle's theorem.

>Verify Rolle's theorem for the function $f(x) =x^2$ in the interval [-1,1]

In answer they didn't change $x$. 

>Verify Rolle's theorem for the function $f(x)=x^{\frac{2}{3}}$ in the interval (-1,1).

They had written $0$ instead of x this time. I am confused how to find the value. In earlier problem it was closed interval. And, it is open interval now. Which law should I apply to find the value?

>For the function $f(x)=x when 0<=x<=1$ $f(x)=2-x when 1<x<=2$ f(0)=0=f(2), according to the Rolle's theorem there exist at least one point in the interval (0,2) such that $f'(x)=0$ but $f'(x)!=0$ for any $x element (0,2)$ why not, explain.

This time they had put $1$ instead of $x$.

calculus real-analysis limits differentiability rolle's-theorem

Communities

Post History