Wemazi

Topic 5.1

Find every matching point

Builds on checking when the theorem applies. 7 minutes.

Two tangents parallel to one secant line

The curve is the graph of \(f(x) = x^3 - 6x\) on \([-3, 3]\). The secant line joins the end points \((-3, -9)\) and \((3, 9)\), and it rises 18 units over 6 units, so its slope is 3.

Drag the orange point from left to right. The orange line is the tangent, and the dashed line through the point has the same slope as the secant line.

The tangent is parallel to the secant line twice, at \(x = -\sqrt3\) and at \(x = \sqrt3\), where the slope of the tangent is 3.

A point between the end points where the tangent is parallel to the secant line is called a matching point. This curve has two matching points.

Solve for every point

The lower graph shows the slope of the tangent at each point. Its dashed line sits at the height of the slope of the secant line.

Drag the point along the curve, also past a and b. The orange bar reaches the dashed line wherever the tangent is parallel to the secant line.

This happens three times, but one of them is left of \(a\), where the curve is faint. Only the two points between \(a\) and \(b\) are matching points, because the theorem promises a point strictly inside the interval.

Notation

  1. Check that \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\). Then find the slope \(m\) of the secant line, which is the average rate of change.

    \[\cy{m} = \dfrac{f(b) - f(a)}{b - a}\]

  2. Differentiate \(f\). Then solve the equation that sets the slope of the tangent equal to \(m\).

    \[f'(c) = \cy{m}\]

  3. Keep a solution only if it lies strictly between \(a\) and \(b\). An end point does not count, because the theorem names a point inside the interval.

    \[a < c < b\]

  4. The answer is every solution that you keep. There can be one solution or several.

When it fails

Move the right end to \(b = 1\). Then \(f(1) = -5\), so the secant line has slope \(\tfrac{-5 + 9}{4} = 1\). The derivative is \(f'(c) = 3c^2 - 6\), so the equation \(3c^2 - 6 = 1\) gives \(c = \pm\sqrt{7/3}\), which is about \(\pm 1.53\).

Only \(c = -\sqrt{7/3}\) lies between \(-3\) and 1. The tangent at \(x = \sqrt{7/3}\) is also parallel to the secant line, but that point is outside the interval, so it is not a matching point.

Example 1: a parabola

Find every point \(c\) in \((1, 7)\) where \(f'(c)\) equals the average rate of change of \(f(x) = x^2 - 6x\) on \([1, 7]\).

  1. The function is a polynomial, so both conditions hold. The end values are \(f(1) = -5\) and \(f(7) = 7\), so the secant line has slope 2.

    \[\cy{m} = \dfrac{7 - (-5)}{7 - 1} = 2\]

  2. The derivative is \(f'(x) = 2x - 6\). Solve \(f'(c) = 2\).

    \[2c - 6 = 2 \quad\Rightarrow\quad c = 4\]

  3. The number 4 lies between 1 and 7, so it is the one matching point.

Example 2: a cubic

Find every matching point for \(g(x) = x^3 - 9x\) on \([0, 4]\).

  1. The end values are \(g(0) = 0\) and \(g(4) = 28\).

    \[g(4) = 64 - 36 = 28\]

  2. A tempting wrong move: The slope of the secant line is the change in \(g\), which is 28.

    \[m = g(4) - g(0) = 28\]

    In the figure, the secant line rises 28 units over 4 units, so it is much flatter than a line of slope 28. A tangent with slope 28 is not parallel to it.

  3. Divide the change in \(g\) by the change in \(x\). The slope of the secant line is 7.

    \[\cy{m} = \dfrac{28 - 0}{4 - 0} = 7\]

  4. The derivative is \(g'(x) = 3x^2 - 9\), so solve \(3c^2 - 9 = 7\). The solution \(c = -\tfrac{4}{\sqrt3}\) is negative, so it is outside \([0, 4]\), and only \(c = \tfrac{4}{\sqrt3} \approx 2.31\) is a matching point.

    \[3c^2 - 9 = 7 \quad\Rightarrow\quad c = \pm\dfrac{4}{\sqrt3}\]

Practice

Find every matching point for \(h(x) = x^3 - 3x^2\) on \([-1, 3]\).

Find the slope of the secant line first, and drag the point until the slope beside the tangent equals it. Solve for c and keep only the solutions between −1 and 3.

Solution

  1. The end values are \(h(-1) = -4\) and \(h(3) = 0\), so the secant line has slope 1.

    \[\cy{m} = \dfrac{0 - (-4)}{3 - (-1)} = 1\]

  2. The derivative is \(h'(x) = 3x^2 - 6x\). Solve \(3c^2 - 6c = 1\) with the quadratic formula.

    \[\begin{aligned} 3c^2 - 6c - 1 &= 0 \\ c &= \dfrac{6 \pm \sqrt{48}}{6} \\ &= 1 \pm \dfrac{2}{\sqrt3} \end{aligned}\]

  3. Both solutions, about −0.15 and 2.15, lie between −1 and 3, so there are two matching points.

Questions

The graph shows a function \(f\) with \(f(0) = 0\), \(f(2) = 5\), \(f(4) = 4\) and \(f(6) = 9\).

  1. What is the slope of the secant line from \(x = 0\) to \(x = 6\)?

    • \(\tfrac32\)
    • 9
    • \(\tfrac23\)
    Answer

    \(\tfrac32\)

  2. How many points between \(x = 2\) and \(x = 6\) have a tangent parallel to the secant line from \(x = 2\) to \(x = 6\)?

    • One
    • Two
    • None
    Answer

    One

  3. Find the matching point for \(f(x) = x^2 + 3x\) on \([1, 5]\).

    • \(c = 3\)
    • \(c = 9\)
    • \(c = 4.5\)
    • \(c = 16.5\)
    Answer

    \(c = 3\)

Summary

You find the slope of the secant line, set \(f'(c)\) equal to that slope, and keep only the solutions strictly between the end points.

This works because at a matching point the tangent is parallel to the secant line, and the slope of the tangent is \(f'(c)\).

You use it when a problem asks for the point or the points that the theorem guarantees.

This is one of five skills in topic 5.1 of Which & Why: AP Calculus.