Differentiability of the mean value

[jostpuur]

So if a function

$$f:[a,b]\to\mathbb{R}$$

is differentiable, then then for each $x\in [a,b]$ there exists $\xi_x \in [a,x]$ so that

$$f'(\xi_x) = \frac{f(x)-f(a)}{x-a}$$

Sometimes there may be several possible choices for $\xi_x$. My question is, that if the mapping $x\mapsto \xi_x$ is chosen so that it is continuous, is it always also differentiable? In other words, does the limit

[tex]
\lim_{h\to 0}\frac{\xi_{x+h}-\xi_x}{h}
[/itex]

exist?

 

[MathematicalPhysicist]

what you wrote is correct when: f'(e)(x-a)=f(x)-f(a), cause of the choice of x=a.
for your question you mean the function at the points ksi_x+h and at ksi_x.

well this is ofcourse correct when f is differentaibale continuous.
btw, it's enough to assume that it's differentiable in (a,b) and continuous in [a,b].

 

[tacman]

Differentiability of the mean value is a fundamental concept in calculus and is closely related to the concept of the derivative of a function. The statement that for a differentiable function f:[a,b]\to\mathbb{R}, there exists a point \xi_x \in [a,x] such that f'(\xi_x) = \frac{f(x)-f(a)}{x-a} is known as the Mean Value Theorem. This theorem plays a crucial role in many applications of calculus, such as optimization and curve sketching.

The question posed in this content is whether the mapping x\mapsto \xi_x, chosen to be continuous, is also differentiable. The answer to this question is yes, the mapping is in fact differentiable. This can be seen by considering the definition of differentiability. A function g is said to be differentiable at a point x_0 if the limit

[tex]
\lim_{h\to 0}\frac{g(x_0+h)-g(x_0)}{h}
[/itex]

exists. In this case, if we let g(x) = \xi_x, then we have

[tex]
\lim_{h\to 0}\frac{\xi_{x+h}-\xi_x}{h} = \lim_{h\to 0}\frac{\frac{f(x+h)-f(a)}{x+h-a}-\frac{f(x)-f(a)}{x-a}}{h} = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h(x+h-a)} = \lim_{h\to 0}\frac{f'(\xi_{x+h})}{x+h-a} = \frac{f'(x)}{x-a}
[/itex]

where we have used the Mean Value Theorem in the third equality. Since f is differentiable, f' is continuous and thus the limit exists. Therefore, the mapping x\mapsto \xi_x is differentiable.

In conclusion, the mapping x\mapsto \xi_x, chosen to be continuous, is also differentiable. This result is an important consequence of the Mean Value Theorem and highlights the connection between differentiability and continuity.