Math Applets

Rolle’s Theorem

How to use the applet

• Drag the green point $a$ along the $x$-axis to choose the left endpoint.
• The red point $b$ is computed automatically so that $f(a)=f(b)$ (here $b=-a$), so the main hypothesis of Rolle’s theorem is always satisfied.
• The dashed vertical segments indicate the values $f(a)$ and $f(b)$.
• The purple points $\xi$ mark the critical points of the chosen function $f(x)=x^4-2x^2$, and a purple tangent is drawn there when $\xi\in(a,b)$.
• As you move $a$, the interval $(a,b)$ changes.
• Whenever at least one of the marked points $\xi$ lies strictly between $a$ and $b$, the tangent at that point is horizontal, illustrating $f'(\xi)=0$.
• For this function, the critical points are $\xi\in\{-1,0,1\}$ (since $f'(x)=4x(x^2-1)$), and the applet shows exactly which of them fall inside $(a,b)$.

Description

Link to code.

Reference: Lecture Notes Calculus 1 ( May22,2022 ) Theorem6.9 page 104

The applet is an attempt to visualize the Rolle’s theorem interactively. The student drag points $a$ and $b$, to change the interval.

When the theorem conditions are met ($f(a) = f(b)$):

const eps = 0.1;

function rolleConditionHolds() {
    return Math.abs(f(pointA.X()) - f(pointB.X())) < eps;
}

The student can see horizontal tangent lines for each critical point $\xi$ inside the interval. Using a symmetric function helps the student create this condition.

Morover the chosen function is simple to deal with for calculting its critical points.

Indeed $f(x) = x^4 - 2x^2$ with $f'(x)=4x^3 - 4x = 4x(x^2 -1)$. Therefore $f'(x)=0$ for $x= 1, -1$ and $x=0$.

const xis = [-1, 0, 1];

xis.forEach((xi) => {

    const pointXiX = createPoint(
        board,
        [xi, 0],
        {
            name: `ξ`,
            visible: () => xiVisible(xi),
        },
        COLORS.purple
    );

    const pointXi = createPoint(
        board,
        [xi, f(xi)],
        {
            name: "f(ξ)",
            visible: () => xiVisible(xi),
        },
        COLORS.purple
    );
...