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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

Mvt2

For any function that is continuous on [a,b] and differentiable on (a,b) there exists some c\in(a,b) such that the secant joining the endpoints of the interval [a,b] is parallel to the tangent at c .

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 svg

Geometrical meaning of Cauchy's 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] .

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