Mean Value Theorem

How to use the applet

• Drag the points $a$ (green) and $b$ (red) along the $x$-axis to change the interval $[a,b]$.
• The points $(a,f(a))$ and $(b,f(b))$ are marked on the graph.
• The red line is the secant through $(a,f(a))$ and $(b,f(b))$. Its slope is $\frac{f(b)-f(a)}{b-a}$.
• The point $c$ (purple) is computed so that $f'(c)$ equals this secant slope.
• The purple line is drawn parallel to the secant through $(c,f(c))$, visualizing that the tangent slope at $c$ matches the secant slope.
• As you move $a$ and $b$, the secant slope changes, and the applet updates $c$ accordingly.
• The theorem guarantees that at least one such point $c\in(a,b)$ exists (for functions continuous on $[a,b]$ and differentiable on $(a,b)$); in this example the applet displays one solution.

Description

Link to code.

Reference: Lecture Notes Calculus 1 ( May22,2022 ) Figure6.6 and Theorem6.10 page 104

To demonstrate the theorem i have selected a simple function $0.2*x^3 - 0.5*x^2 + 2$ with derivative $0.6x^2 - x$.

    const f = (x: number) => 0.2 * x * x * x - 0.5 * x * x + 2;
    // const fPrime = (x: number) => 0.6 * x * x - x;

Points $a$ and $b$ can be adjusted. The secant is just a line passing though $f(a)$ and $f(b)$. To construct $c$ i solved the equation:

Giving $c^2 - c - \frac{f(b) - f(a)}{(b-a)*0.6}$. Solving with the quadratric formula: $c =1 ± \frac{\sqrt{1 + 4*0.6*\text{secantSlope}}}{2*0.6}$. If the discriminant is negative or $c$ does not lie on the interval $[a,b]$ i return NaN (not a number).

    function findC(): number {
        const a = pointA.X();
        const b = pointB.X();
        const fa = f(a);
        const fb = f(b);
        const secantSlope = (fb - fa) / (b - a);
        const discriminant = 1 + 4 * 0.6 * secantSlope;
        if (discriminant < 0) {
            return NaN;
        }
        const cPos = (1 + Math.sqrt(discriminant)) / (2 * 0.6);
        const cNeg = (1 - Math.sqrt(discriminant)) / (2 * 0.6);
        const minAB = Math.min(a, b);
        const maxAB = Math.max(a, b);
        if (cPos >= minAB && cPos <= maxAB) {
            return cPos;
        } else if (cNeg >= minAB && cNeg <= maxAB) {
            return cNeg;
        } else {
            return NaN;
        }
    }

Finally the tangent line (best approximation of $f'(c)$) has to be parallel to the secant.

    board.create('parallel', [secant, pointC],...)