I'm trying to understand a certain step within the proof for the mean value theorem, which states that if $f : [a,b] \rightarrow \mathbb{R}$ is differentiable, there exists an $x_0 \in [a,b]$ with $\frac{f(b)-f(a)}{a-b} = f^{\prime}$.

For the proof, we first prove that if a differentiable function $f : [a,b] \rightarrow \mathbb{R}$ has a maximum or minimum in an inner point of $[a,b]$, it's derivative in that point is equal to zero. In order to see that, let's suppose that $x_0 \in [a,b]$ is the x-coordinate for the maximum point, then we have: $\forall x \in [a,b]: f(x)-f(x_0) \leq 0$ and therefore:

for $x<x_0: \frac{f(x)-f(x_0)}{x-x_0} \geq 0$ and also

for $x>x_0: \frac{f(x)-f(x_0)}{x-x_0} \leq 0$.

Because of that, we have:

limit from the left: $lim_{x\nearrow x_0} \frac{f(x)-f(x_0)}{x-x_0} \geq 0$

limit from the right: $lim_{x\searrow x_0} \frac{f(x)-f(x_0)}{x-x_0} \leq 0$

and since we know that for a differentiable function those limits are equal, the derivative in that point will be zero: $lim_{x\rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0} = 0$ $\square$.

Now, to prove the mean value theorem, we define:

$g(x) = f(x) - secant(x)= f(x) - (f(a) + \frac{f(b)-f(a)}{b-a}(x-a))$, which fulfills:

$g(a) = 0, g(b) = 0$

If $g(x)$ now is constant, $g^{\prime}(x)=0 \Leftrightarrow \frac{f(b)-f(a)}{a-b} = f^{\prime}(x)$ If $g(x)$ is not constant, it has a maximum or minimum in in an inner point of $x_0$ on $[a,b]$, where we have $g^{\prime}(x) = 0 \Leftrightarrow \frac{f(b)-f(a)}{a-b} = f^{\prime}(x) \square$.

Now, I understand the whole proof except for the last two lines, in which I can't get my head wrapped around why if the function that is definded by "function - secant" isn't increasing or decreasing, the derivative in that point must be equal to the secant.

Thanks for anyone reading that super long proof and excuse any inaccuracies in the proof since I tried translating it from my native language. Please point them out. Any tips for my problem are appreciated.


$g$ is a differentiable function on $[a, b]$, so we can use theorem about derivative in max/min been zero.

If $g$ isn't constant, then it has either maximum or minimum (or both) in inner point. Indeed, if $g$ isn't constant, then $g(x) \neq 0$ for some $x$. Wlog we can assume $g(x) > 0$. Now take $y_0 = \sup \{g(x) | x \in [a, b]\}$ - as $y_0 > g(x)$, we have $y_0 > 0$. As $g$ is continuous, for some $x_0$ we have $g(x_0) = y_0$. And as $g(a) = g(b) = 0$, we have $x_0 \neq a$, $x_0 \neq b$, so $x_0$ is an inner point.

From first theorem, $g'(x_0) = 0$. Now substitute expression for $g$: $$g(x) = f(x) - (f(a) + \frac{f(b) - f(a)}{b - a}(x - a))$$ $$g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a}$$ $$0 = g'(x_0) = f'(x_0) - \frac{f(b) - f(a)}{b - a}$$ $$f'(x_0) = \frac{f(b) - f(a)}{b - a}$$

It's a specific case of a bit more general construction: if function $h(x) = f(x) - cx$ has an extremum (and thus zero derivative) in $x_0$, you have $f'(x_0) = c$. You can think about adding $-cx$ to function as rotating it's graph around origin. And if rotated graph has horizontal tangent at some point, then original graph had tangent with angular coefficient $c$ at it.

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  • $\begingroup$ Thanks, very good answer. $\endgroup$ – psyph Aug 21 '19 at 21:23

The value of $|g(x)|$ describes the height of the green vertical line at a specific $x$-coordinate

enter image description here $$g(x) = f(x)-\text{secant}(x)$$ According to the Rolle's theorem and to the fact, that $g(a)=g(b)=0$, which is an assumption of the rolle's theorem, we conclude that there is a coordinate $c$, where $g'(c)=0$. With the help of this last equation and with the help of the equation of the secant we get to the conclusion of the mean value theorem. We can get the formula of the secant line $\text{secant}(x)$ from the fact, that $$\frac{\text{secant}(x)-f(a)}{x-a}=\frac{f(b)-f(a)}{b-a}$$

Which for example can be derived with the help of a property of similar triangles, this property is: the side lengths of two similar triangles are proportional. For the proof, look here. Also look here.

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    $\begingroup$ Thanks for helpful your answer. $\endgroup$ – psyph Aug 21 '19 at 21:23

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