Post History
#3: Post edited
- 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 $x$.
- <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 $f'(x) \neq 0$ for any $x \in (0,2)$ 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
- 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 $x$.
- 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 $x$.
- <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
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 $x$.