The mean value theorem states that in a closed interval, a function has at least one point where the slope of a tangent line at that point (i.e. the derivative) is equal to the average slope of the function (or the secant line between the two endpoints).

Ergo:

${\displaystyle f(x)}$ on a closed interval ${\displaystyle [a, b]}$ has a derivative at point ${\displaystyle c}$ , which has an equivalent slope to the one connecting ${\displaystyle a}$ and ${\displaystyle b}$ .

Therefore, the derivative equals the slope formula:

${\displaystyle f'(c) = \frac{f(b)-f(a)}{(b-a)}}$

There are three formulations of the mean value theorem:

Rolle's Theorem

Rolle's theorem states that for a function ${\displaystyle f:[a,b]\to\R}$ that is continuous on ${\displaystyle [a, b]}$ and differentiable on ${\displaystyle (a,b)}$:

If ${\displaystyle f(a) = f(b)}$ then ${\displaystyle \exists c\in(a,b):f'(c)=0}$

Proof

By the Weierstrass Theorem, the function ${\displaystyle f}$ has two extrema in ${\displaystyle [a, b]}$ , say a minimum ${\displaystyle x_1}$ and a maximum ${\displaystyle x_2}$ . There are two cases:

(i) If ${\displaystyle x_1,x_2\in\{a,b\}}$ then ${\displaystyle f(x_1) = f(x_2)}$ (by the condition for Rolle's Theorem to hold). However, ${\displaystyle \forall x\in[a,b],f(x)\ge f(x_1)}$ (as it's a minimum) and ${\displaystyle f(x)\le f(x_2)}$ (as it's a maximum). So ${\displaystyle f(x_1)\le f(x)\le f(x_2)=f(x_1)\implies f(x)=f(x_1),\forall x\in[a,b]}$ .
Therefore ${\displaystyle f}$ is constant on ${\displaystyle [a, b]}$ , so its derivative is 0 everywhere, so there certainly exists a ${\displaystyle c \in (a, b)}$ with ${\displaystyle f'(c)=0}$ .

(ii) If the above case does not happen, then ${\displaystyle \exists i\in\{1,2\}:x_i\in(a,b)}$ . So take ${\displaystyle c=x_i}$ , and as it is an extremum ${\displaystyle f'(c)=0}$ .

Lagrange's Mean Value Theorem

Lagrange's mean value theorem, sometimes just called the mean value theorem, states that for a function ${\displaystyle f:[a,b]\to\R}$ that is continuous on ${\displaystyle [a, b]}$ and differentiable on ${\displaystyle (a,b)}$:

${\displaystyle \exists c\in(a,b):f'(c)=\frac{f(b)-f(a)}{b-a}}$

Proof

Rather than prove this theorem explicitly, it is possible to show that it follows directly from Rolle's theorem. As we have already proved Rolle's, this is enough.

Define a function ${\displaystyle h(x)=f(x)-\left(f(a)+\frac{f(b)-f(a)}{b-a}(x-a)\right)}$

Observe

${\displaystyle h(a)=f(a)-\left(f(a)+\frac{f(b)-f(a)}{b-a}(a-a)\right)=0}$

And

${\displaystyle h(b)=f(b)-\left(f(a)+\frac{f(b)-f(a)}{b-a}(b-a)\right)=0}$

Note that by the algebra of continuous and differentiable functions, ${\displaystyle h}$ satisfies the conditions for Rolle's Theorem.

So by the theorem, ${\displaystyle \exists c\in(a,b):h'(c)=0}$ ,

So ${\displaystyle h'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}=0}$ ,

i.e. ${\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}}$ .

Note also that Rolle's Theorem is a special case of Lagrange's MVT, where ${\displaystyle f'(c)=0}$ .

Cauchy's Mean Value Theorem

Cauchy's mean value theorem states that for two functions ${\displaystyle f,g:[a,b]\to\R}$ that are continuous on ${\displaystyle [a, b]}$ and differentiable on ${\displaystyle (a,b)}$ :

${\displaystyle \exists c\in(a,b):\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\quad, g(a)\ne g(b)}$

Proof

Note: ${\displaystyle g(a) \ne g(b)}$ because saying the opposite will apply Rolle's Theorem that ${\displaystyle \exists x\in(a,b):g'(x)=0}$ .

Again we can show this follows from Rolle's Theorem:

Define a function ${\displaystyle h(x)=f(x)-\left(f(a)+\frac{f(b)-f(a)}{g(b)-g(a)}\big(g(x)-g(a)\big)\right)}$

Observe

${\displaystyle h(a)=f(a)-\left(f(a)+\frac{f(b)-f(a)}{g(b)-g(a)}\big(g(a)-g(a)\big)\right)=0}$

And

${\displaystyle h(b)=f(b)-\left(f(a)+\frac{f(b)-f(a)}{g(b)-g(a)}\big(g(b)-g(a)\big)\right)=0}$

Again, by the algebra of continuous and differentiable functions, ${\displaystyle h}$ also satisfies the other conditions for Rolle's Theorem.

So by the theorem, ${\displaystyle \exists c\in(a,b):h'(c)=0}$

So ${\displaystyle h'(c)=f'(c)-\frac{f(b)-f(a)}{g(b)-g(a)}g'(c)=0}$

i.e. ${\displaystyle \frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\qquad,\forall x\in(a,b):g'(x)\ne0}$ .

Note that Lagrange's MVT (and therefore also Rolle's Theorem) is just a special case of Cauchy's MVT, where you take ${\displaystyle g(x)=x,\forall x\in[a,b]}$ .