Cauchy’s mean value theorem

Example 1, from math.stackexchange

How to use the applet:

• Drag the blue points $a$ and $b$ along the $x$-axis to choose the interval $[a,b]$.
• Drag the purple point $\xi$ along the $x$-axis (it is constrained to stay inside $[a,b]$).
• The blue curve is $f(x)$.
• The orange curve is $r\,g(x)$, where$r=\frac{f(b)-f(a)}{g(b)-g(a)}.$
• The red curve (in the lower panel) is $h(x)=f(x)-r\,g(x),$ constructed so that $h(a)=h(b)$.

1. First choose $a$ and $b$. The applet automatically recomputes $r$ so that the endpoints of $h$ match.
2. Now move $\xi$ and look at the tangents at $x=\xi$.
   • When you hit a point where $h'(\xi)\approx 0$, the applet highlights $\xi$ (turns green).
3. At such a $\xi$, the tangents to $f$ and $r\,g$ have the same slope, because

Example 2, extension of MVT as parametric curve

How to use

Instead of drawing $f$ and $g$ as graphs in the $(x,y)$-plane, we look at the parametric curve $\gamma(t)=(g(t),\,f(t))$ in the $(g,f)$-plane. In this applet $g(t)=t$, so the horizontal coordinate is just $t$.

• Drag the green point $A=(g(a),f(a))$ and the red point $B=(g(b),f(b))$ along the blue curve $\gamma$. (They determine the parameters $a$ and $b$, i.e. the interval $[a,b]$.)
• The orange line is the secant through $A$ and $B$ in the $(g,f)$-plane. Its slope is $\frac{f(b)-f(a)}{g(b)-g(a)}.$
• The purple point is $\gamma(\xi)=(g(\xi),f(\xi))$, and the purple line is drawn parallel to the secant through that point.
• CMVT guarantees that there exists $\xi\in(a,b)$ such that

• In the $(g,f)$-plane, the derivative ratio $\frac{f'(\xi)}{g'(\xi)}$ is exactly the slope of the tangent to $\gamma$ at $\xi$ (because $\gamma'(t)=(g'(t),f'(t))$). So “parallel to the secant” is the geometric way to see the theorem.
• The applet computes one such $\xi$ automatically (so you can focus on the geometry rather than solving the equation by hand).

Description

Reference: Lecture Notes Calculus 1 ( May22,2022 ) Lemma6.15 page 109 The applet are an attempt to show the Cauchy’s mean value theorem.

Example 1

Link to code.

This example is taken from this Math StackExchange answer.

The argument is closed to the proof used in the lecture notes, and provides a nice geometric interpretation.

Given an interval $[a,b]$, the applet displays the functions $f(x)$, $r\,g(x)$, and

The choice of $r$ is designed so that the endpoints match:

Therefore, by Rolle’s Theorem, there exists $\xi\in(a,b)$ such that

Since $r$ is constant once $a$ and $b$ are fixed, the derivative is

computed analytically as a combination of $f'$ and $g'$.

const getR = () => {
    const a = getA();
    const b = getB();
    const denom = g(b) - g(a);
    if (Math.abs(denom) < 1e-10) return NaN;
    return (f(b) - f(a)) / denom;
};

// h'(x) = f'(x) - r g'(x)
const dh = (x: number) => df(x) - getR() * dg(x);

After fixing $a$ and $b$, the student can move the point $\xi$ along the interval and try to satisfy the condition $h'(\xi)\approx 0$.
When this happens, the applet highlights the geometric meaning of the theorem: at $x=\xi$ the tangents to $f$ and $r\,g$ have the same slope, which corresponds to the CMVT conclusion

Example 2

Link to code. This applet visualizes Cauchy’s Mean Value Theorem (CMVT) through a geometric interpretation in the $(g,f)$-plane.

Instead of drawing the graphs of $f$ and $g$ in the usual $(x,y)$-plane, we consider the parametric curve

const curve = board.create(
  "curve",
  [(t: number) => g(t), (t: number) => f(t), xmin, xmax],
  { strokeColor: COLORS.blue, strokeWidth: 4 }
) as JXG.Curve;

We choose a particularly simple function

so that:

• $g'(t)=1\neq 0$ everywhere),
• $g^{-1}(u)=u$ is known explicitly,
• the parameter $t$ is exactly the x–coordinate of the point on the curve.

Two movable points $(g(a),f(a))$ (green) and $(g(b),f(b))$ (red). The applet draws the secant line through $(g(a),f(a))$ and $(g(b),f(b))$.

In this applet we take:

Let

Then

Expand:

To find $\xi$, we solve $h'(t)=0$:

This is why in the code findXi() computes:

• the interval endpoints in parameter space:
  • left = min(a,b), right = max(a,b)
• dF = f(right) - f(left)
• dG = right - left
• then

as t2 = (dF + dG) / (0.9 * dG).

Since $t^2=t2$, the candidates are

The code then:

1. discards the case $t2<0$ (no real square root; numerically this can occur for some endpoint choices),
2. checks which candidate lies inside left, right,
3. returns that one (if both are inside, it picks the one closer to the midpoint).

Finally the tangnet at $\xi$ should be parallel to the secant:

// Secant line AB
const secant = board.create("line", [A, B], {
  strokeColor: COLORS.orange,
  strokeWidth: 3,
}) as JXG.Line;

// ξ point on the parametric curve
const Xi = board.create(
  "point",
  [() => g(findXi()), () => f(findXi())],
  { name: "", size: 5, fixed: true, strokeColor: COLORS.purple, fillColor: COLORS.purple }
) as JXG.Point;

board.create("parallel", [secant, Xi], {
  strokeColor: COLORS.purple,
  strokeWidth: 3,
});