Distinguish existence from uniqueness

One secant line, a changing curve

The curve starts at (0, 0) and ends at (4, 4). The straight line that joins its ends is the secant line, and its slope is 1. Each orange label c marks a matching point, where the tangent is parallel to the secant line.

With the slider at 0, the curve is \(f(x) = 3x - \tfrac12 x^2\), and the tangent at \(x = 2\) is the one matching point.

Move the slider to add a wiggle. The end points stay fixed, so the secant line does not change.

Past a wiggle of 0.25, a second matching point enters the interval at its right end. At 0.5 there are two, at about \(x = 1.13\) and \(x = 3.54\).

The theorem promises at least one matching point. It does not promise only one.

At least one

As the curve changes, at least one orange label stays on it, but the number of labels varies.

The theorem guarantees at least one matching point, and the curve decides how many.

Notation

1. The theorem is a statement of existence. If \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\), it promises at least one number \(c\) in \((a, b)\) with this property.

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

2. To claim exactly one such \(c\), count the solutions of \(f'(c) = m\) in \((a, b)\). The dashed line on the lower graph is at the secant slope 1. With a large wiggle, \(f'\) crosses it twice, so \(f'(c) = 1\) has two solutions in \((0, 4)\).

3. If the curve bends one way along the whole interval, the tangent turns in one direction, so its slope reaches the secant slope once. A curve with a wiggle bends both ways, so its slope can reach that value twice.

When it fails

Add the wiggle. Then \(f'(x) = \tfrac32 x^2 - 7x + 7\), and the graph of \(f'\) falls and then rises.

It crosses the dashed line twice, so there are two matching points. A guarantee of at least one point does not rule out more.

Example 1: a cube

Show that \(f(x) = x^3\) has exactly one point \(c\) in \((1, 3)\) where \(f'(c)\) equals the average rate of change on \([1, 3]\).

1. The function is a polynomial, so the theorem applies. The secant line has slope 13.

   \[\cy{m} = \dfrac{27 - 1}{3 - 1} = 13\]

2. Set \(f'(c) = 3c^2\) equal to 13. Of the solutions \(\pm\sqrt{13/3}\), only the positive one lies in \((1, 3)\).

   \[c = \sqrt{\dfrac{13}{3}} \approx 2.08\]

3. On the lower graph, \(f'\) crosses the dashed line at 13 once between 1 and 3, so there is exactly one matching point.

Example 2: counting solutions

Show that the equation \(x^3 + 2x - 5 = 0\) has exactly one real solution.

1. Let \(f(x) = x^3 + 2x - 5\). It is continuous, and its values at 1 and 2 have opposite signs, so the graph crosses the \(x\)-axis between them.

   \[f(1) = -2, \quad f(2) = 7\]

2. A tempting wrong move: The sign change shows that there is exactly one solution.

   A sign change shows at least one crossing, not how many. The curve now shown has the same two values and crosses the axis three times.

3. Suppose there were two solutions, \(a < b\). Then \(f(a) = f(b) = 0\), so the secant line from \(a\) to \(b\) is horizontal. Because \(f\) is a polynomial, the theorem gives a point \(c\) between them with \(f'(c) = 0\), a horizontal tangent. The figure shows this on the curve with three crossings.

4. But the derivative is at least 2 for every \(x\), so no tangent is horizontal. Two solutions would force a horizontal tangent, so there is only one solution.

   \[f'(x) = 3x^2 + 2 \ge 2\]

Practice

Find the point \(c\) in \((0, 2)\) where \(f'(c)\) equals the average rate of change of \(f(x) = x^4\) on \([0, 2]\). Then explain why it is the only one.

Drag the point until the tangent is parallel to the secant line. Can that happen anywhere else?

Questions

The graph shows a function \(f\) that is continuous on \([0, 6]\) and differentiable on \((0, 6)\), with \(f(0) = f(6) = 2\). It has two peaks, at \(x = 1.5\) and \(x = 4.5\), and a valley at \(x = 3\).

1. How many points between \(x = 0\) and \(x = 6\) have a horizontal tangent?

   • 3
   • 1
   • 2
   Answer

   3

2. The secant line from the peak at \(x = 1.5\) to the valley at \(x = 3\) has slope \(-2\). How many points between 1.5 and 3 have a tangent with slope \(-2\)?

   • 2
   • 1
   • 0
   Answer

   2

3. For \(f(x) = 2x^3 - 5x\) on \([-2, 4]\), how many points \(c\) in \((-2, 4)\) have \(f'(c)\) equal to the average rate of change?

   • Exactly one
   • Exactly two
   • At least one, but the number cannot be found
   • None
   Answer

   Exactly one

Summary

You read the theorem as a promise of at least one matching point, and you claim exactly one only when you have a reason.

The tangent slope can equal the secant slope more than once, so the point is unique only if \(f'(c) = m\) has one solution in \((a, b)\).

You use it when a problem asks how many matching points there are, or whether the point is the only one.

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