jostpuur
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- TL;DR Summary
- Can the points produced by Mean Value Theorem be chosen to be continuous?
Suppose f:]a,b[\to\mathbb{R} is some differentiable function. Then it is possible to define a new function
<br /> ]a,b[\to [a,b],\quad x\mapsto \xi_x<br />
in such way that
<br /> f(x) - f(a) = (x - a)f'(\xi_x)<br />
for all x\in ]a,b[. Mean Value Theorem says that these \xi_x exist.
One question that sometimes interests us is that is the function x\mapsto \xi_x continuous. The simple answer is that no, this function is not necessarily continuous.
However, for any fixed x, there can exist many different \xi_x with the same mean value property. So Mean Value Theorem doesn't fix the function x\mapsto \xi_x uniquely; it only says that at least one this kind of function exists.
So the big question that arises is that is it possible to choose the values \xi_x in a such way that the function x\mapsto\xi_x becomes continuous?
<br /> ]a,b[\to [a,b],\quad x\mapsto \xi_x<br />
in such way that
<br /> f(x) - f(a) = (x - a)f'(\xi_x)<br />
for all x\in ]a,b[. Mean Value Theorem says that these \xi_x exist.
One question that sometimes interests us is that is the function x\mapsto \xi_x continuous. The simple answer is that no, this function is not necessarily continuous.
However, for any fixed x, there can exist many different \xi_x with the same mean value property. So Mean Value Theorem doesn't fix the function x\mapsto \xi_x uniquely; it only says that at least one this kind of function exists.
So the big question that arises is that is it possible to choose the values \xi_x in a such way that the function x\mapsto\xi_x becomes continuous?