FANDOM

1,025 Pages

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:

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

Therefore, the derivative equals the slope formula:

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

There are three formulations of the mean value theorem:

Rolle's Theorem

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

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

Proof

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

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

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

Lagrange's Mean Value Theorem

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

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

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

Observe

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

And

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

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

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

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

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

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

Cauchy's Mean Value Theorem

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

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

Proof

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

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

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

Observe

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

And

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

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

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

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

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

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