Extended mean value theorem

The extended mean value theorem (also called Cauchy's mean value theorem) is usually formulated as:

Let

[math]\displaystyle{ f, g: [a,b] \to \mathbb{R} }[/math]

be continuous functions that are differentiable on the open interval [math]\displaystyle{ (a,b) }[/math]. If [math]\displaystyle{ g'(x)\neq 0 }[/math] for all [math]\displaystyle{ x\in(a,b) }[/math], then there exists a value [math]\displaystyle{ \xi \in (a,b) }[/math] such that

[math]\displaystyle{ \frac{f'(\xi)}{g'(\xi)} = \frac{f(b)-f(a)}{g(b)-g(b)}. }[/math]

Remark: It seems to be easier to state the extended mean value theorem in the following form:

Let

[math]\displaystyle{ f, g: [a,b] \to \mathbb{R} }[/math]

be continuous functions that are differentiable on the open interval [math]\displaystyle{ (a,b) }[/math]. Then there exists a value [math]\displaystyle{ \xi \in (a,b) }[/math] such that

[math]\displaystyle{ f'(\xi)\cdot (g(b)-g(a)) = g'(\xi) \cdot (f(b)-f(b)). }[/math]

This second formulation avoids the need that [math]\displaystyle{ g'(x)\neq 0 }[/math] for all [math]\displaystyle{ x\in(a,b) }[/math] and is therefore much easier to handle numerically.

The proof is similar, just use the function

[math]\displaystyle{ h(x) = f(x)\cdot(g(b)-g(a)) - (g(x)-g(a))\cdot(f(b)-f(a)) }[/math]

and apply Rolle's theorem.

The underlying JavaScript code

var board = JXG.JSXGraph.initBoard('box', {boundingbox: [-5, 10, 7, -6], axis:true});
var p = [];

p[0] = board.create('point', [0, -2], {size:2});
p[1] = board.create('point', [-1.5, 5], {size:2});
p[2] = board.create('point', [1, 4], {size:2});
p[3] = board.create('point', [3, 3], {size:2});

// Curve
var fg = JXG.Math.Numerics.Neville(p);
var graph = board.create('curve', fg, {strokeWidth:3, strokeOpacity:0.5});

// Secant 
line = board.create('line', [p[0], p[3]], {strokeColor:'#ff0000', dash:1});

var df = JXG.Math.Numerics.D(fg[0]);
var dg = JXG.Math.Numerics.D(fg[1]);

// Usually, the extended mean value theorem is formulated as
// df(t) / dg(t) == (p[3].X() - p[0].X()) / (p[3].Y() - p[0].Y())
// We can avoid division by zero with that formulation:
var quot = function(t) {
    return df(t) * (p[3].Y() - p[0].Y()) - dg(t) * (p[3].X() - p[0].X());
};

var r = board.create('glider', [
            function() { return fg[0](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)); },
            function() { return fg[1](JXG.Math.Numerics.root(quot, (fg[3]() + fg[2]) * 0.5)); },
            graph], {name: '', size: 4, fixed:true, color: 'blue'});

board.create('tangent', [r], {strokeColor:'#ff0000'});