Doubt in Lagrange’s Mean Value Theorem

derivativesreal-analysis

Consider f(x), a differentiable funtion from $(a,b) \to R$.

By Lagrange's Mean Value Theorem (LMVT), there exists some c $\in (a,b)$ such that $f(x)-f(y)=f’(c)(x-y)$ for $x,y \in (a,b)$, where $f'(x)$ denotes the derivative of $f(x)$

My question:
As $x$ and $y$ vary over the interval $(a,b)$, is it right to say that $f'(c)$ covers all values of $f'(x)$ in $(a,b)$ (forif $x$ and $y$ are variable, so is $c$)

I am not able to prove or disprove the claim, only able to intuitively convince myself that it is most likely true. Is it? Please help me figure this out, thanks!

P.S.
Hints and mathematical rigour would be appreciated.

Best Answer

By Lagrange's Mean Value Theorem (LMVT), there exists some c $\in (a,b)$ such that $f(x)-f(y)=f’(c)(x-y)$ for $x,y \in (a,b)$, where $f'(x)$ denotes the derivative of $f(x)$.

No. That is NOT what the mean value theorem says. The mean value theorem says

If $f$ is differentiable on $(a,b)$ and continuous on $[a,b]$, then there is a $c \in (a,b)$ such that $f(b) - f(a) = f'(c)(b - a)$.

A slightly weaker version that is provable from it says

If $f$ is differentiable on $(a,b)$ and $x, y \in (a,b)$ with $x \ne y$, then there is a $c$ between $x$ and $y$ with $f(x) - f(y) = f'(c)(x - y)$.

Note the difference between this version and yours: You pick $c$ first, and then demand that it hold for all $x, y$. This is obviously not going to work for non-linear functions. In the true result, the value of $c$ depends on the values $x$ and $y$. Changing $x$ and/or $y$ may require a different value of $c$.

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