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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

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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