In my textbook, there is an alternative perspective on the mean value theorem that I don't understand. When we introduced the mean value theorem the first time, the statement looked like this:


Now, the author wants to introduce the mean value theorem in the n-dimensional space, and he begins like this:

The mean value theorem of a function of one variable can be written like this: Assume $f: I \rightarrow \Bbb R$ is differentiable on I $\subset \Bbb R$ and $x, x + \xi \in I$, then there is a $\lambda \in [0, 1]$ such that

$$f(x + \xi) - f(x) = f'(x + \lambda \xi) * \xi$$

He doesn't give a proof on this, and there are a few things that I don't comprehend.

  1. Why is there a $\xi$ on the right side of the equation? (outside of the parenthesis) What does this give me?

  2. What does $f(x + \xi) - f(x)$ tell me? Shall I comprehend this as the slope of the secant that connects $f(x + \xi)$ and $f(x)$?


The quantity $\xi$ can be understood as $\Delta x$ or as $h$ or as $b-a$; for some reason your author likes $\xi$ instead.

The quantity $f(x+\xi)-f(x)$ is the numerator of the slope of the secant line. The slope itself is $\frac{f(x+\xi)-f(x)}{\xi}$.

As said in the other answer, the $\lambda$ outside the parenthesis is a typo for $\xi$. So your equation, with the typo corrected, is equivalent to $$\frac{f(x+\xi)-f(x)}{\xi} = f'(x+\lambda \xi) \quad\text{for some $\lambda \in [0,1]$} $$ You can also, if you like, replace $x$ by $a$ and $x+\xi$ by $b$, to make the notation match what is in the link you provided, but I'll leave that to you.

All that remains is to understand that as $\lambda$ varies from $0$ to $1$, the expression $x+\lambda\xi$ varies between $x$ and $x+\xi$. Thus, the phrase "$f'(x+\lambda\xi)$ for some $\lambda \in [0,1]$" is equivalent to the phrase "$f'(c)$ for some $c$ between $x$ and $x+\xi$".

  • $\begingroup$ Thank you very much! But I think you made a little mistake in your answer: You wrote "the expression $x + \lambda \xi$ varies between $x$ to $\xi$", but I guess you actually meant "the expression $x + \lambda \xi$ varies between $x$ to $x + \xi$", didn't you? $\endgroup$ – Julian Jun 23 '16 at 13:49
  • $\begingroup$ Yes, thanks for the correction. $\endgroup$ – Lee Mosher Jun 23 '16 at 13:49

This is a typo, $\lambda$ should be $\xi$.

As a counter example, take $x=0,\xi=1,f(x)=\sqrt x$. Then

$$f(1)-f(0)=1=\frac1{2\sqrt{\lambda}}\lambda$$ doesn't work, while

$$f(1)-f(0)=1=\frac1{2\sqrt{\lambda}}$$ does.


$$x\leftrightarrow a,\\x+\xi\leftrightarrow b,\\x+\lambda\xi\leftrightarrow c.$$

  • $\begingroup$ Whoops, that was my own mistake, sorry. I edited the question. $\endgroup$ – Julian Jun 23 '16 at 12:39

What the mean value theorem basically states is that on a continuous function, given two points $(x_1, y_1)$ and $(x_2, y_2)$ where $x_2 > x_1$, the function's derivative somewhere on the interval from $x_1$ to $x_2$ must be equal to the function's average slope $\frac{\Delta y}{\Delta x}$ (or equivalently, $\frac{y_2-y_1}{x_2-x_1}$).

The original equation you gave was $$f(x + \xi) - f(x) = f'(x + \lambda \xi) * \xi$$ which can be rewritten as $$\frac{f(x + \xi) - f(x)}{\xi} = f'(x + \lambda \xi)$$ If we say that $x_1 = x$ and $x_2 = x+\xi$, then the above equation can be rewritten as $$\frac{f(x_2) - f(x_1)}{x_2-x_1}=f'(x+\lambda\xi)$$ Additionally, we can let $x_3 = x+\lambda\xi$. Since $0 < \lambda < 1$, then $x_1 < x_3 < x_2$. Now it is mush easier to see how this mathematical statement is equivalent to the original theorem: $\frac{y_2-y_1}{x_2-x_1} = f'(x_3)$, where $x_3$ is some value between $x_1$ and $x_2$.


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