Choose the strongest justified conclusion

How many times does the line cross?

The curve is the graph of \(f(x) = x^3 - 4x\) from \(x = -3\) to \(x = 3\), which starts at height \(-15\) and ends at height 15.

The orange line is at height \(k = 0\). Because 0 is strictly between \(-15\) and 15 and the curve is unbroken, the line must cross the curve. It crosses at \(x = -2\), \(x = 0\) and \(x = 2\).

Press Set k = 10. The line now crosses the curve only once, near \(x = 2.76\).

Both heights are strictly between the end heights, but one gives three crossings and the other gives one. The only fact true for both is that the line crosses the curve at least once.

At least one, and nothing more is promised

The curve goes from \((a, f(a))\) to \((b, f(b))\) without a break, and the orange line is at a height \(\cy{k}\) strictly between the two end heights.

Drag the slider to bend the middle of the curve. The ends stay fixed, but the number of crossings changes.

With no bend, the curve is straight and crosses the line once. With a large bend, it crosses three times.

In every case the line crosses at least once. The theorem promises this and nothing more.

Notation

1. Suppose \(f\) is continuous on \([a, b]\) and \(\cy{k}\) is strictly between \(\cy{f(a)}\) and \(\cy{f(b)}\). Then there is at least one number \(\cy{c}\) in the open interval \((a, b)\) with \(f(c) = k\).

   \[\exists\, c \in (a, b) : f(c) = k\]

2. The theorem does not say where c is, so name the interval (a, b) that holds c, not a point.

3. The theorem does not say how many numbers c there are. When the curve bends, the red rings mark two more crossings, so the words exactly one claim more than the theorem supports.

4. To claim exactly one number \(c\), add a second fact. If \(f\) is increasing on \((a, b)\), it takes each value at most once. The curve is green where it rises.

   \[\begin{aligned} f \text{ increasing on } (a, b) &\Rightarrow \\ \text{at most one } c & \end{aligned}\]

When it fails

Suppose you claim exactly one number \(c\) with \(f(c) = k\). If the graph is straight, the claim holds, because the orange line crosses it once.

Press Bend the curve. The claim fails, because the orange line now meets the curve three times. The end heights are the same in both pictures, so they cannot give the count.

Example 1: at least one, then exactly one

Show that \(g(x) = x^3 + 2x - 7\) has a zero in \((1, 2)\). Then decide how many zeros \(g\) has on \([1, 2]\).

1. The function g is a polynomial, so it is continuous. Compute the end heights.

   \[g(1) = \cy{-4} \qquad g(2) = \cy{5}\]

2. The number 0 is strictly between \(-4\) and 5, so the theorem gives at least one \(c\) in \((1, 2)\) with \(g(c) = 0\).

   \[-4 < \cy{0} < 5\]

3. The terms \(x^3\) and \(2x\) both increase as \(x\) increases, so their sum \(g\) increases and takes each value at most once. With this second fact, \(g\) has exactly one zero on \([1, 2]\). The curve is green because it rises everywhere.

   \[x_1 < x_2 \Rightarrow g(x_1) < g(x_2)\]

Example 2: a curve that bends back

Use the theorem to describe the zeros of \(h(x) = x^3 - 6x^2 + 9x - 2\) on \([0, 5]\).

1. The function \(h\) is a polynomial, so it is continuous, and 0 is strictly between the end heights.

   \[h(0) = -2 < \cy{0} < 18 = h(5)\]

2. A tempting wrong move: The theorem gives exactly one zero in (0, 5).

   \[\cr{\text{exactly one zero}}\]

   The orange dots show that the curve crosses the x-axis three times, so the claim is false. The theorem gives at least one zero, not a count.

3. To find more zeros, apply the theorem on (0, 1), (1, 3) and (3, 5), because h changes sign on each. These intervals do not overlap, so the three zeros are different.

   \[\begin{aligned} h(0) < 0 &< h(1) \\ h(1) > 0 &> h(3) \\ h(3) < 0 &< h(5) \end{aligned}\]

   A cubic has at most three zeros, so h has exactly three.

Practice

A function \(p\) is continuous on \([1, 6]\), with \(p(1) = -2\) and \(p(6) = 8\). State the strongest conclusion about \(p(x) = 3\).

Drag the bend slider. Every curve passes through the same two end dots, but they do not all reach height 3 the same number of times.

Questions

The graph shows \(r(x) = \tfrac12 x (x - 2)(x - 5)\) on \([-1, 6]\). The orange dots mark the ends.

1. The curve crosses the x-axis three times. What does the theorem conclude for k = 0?

   • There is at least one c in (−1, 6) with r(c) = 0
   • There are exactly three numbers c with r(c) = 0
   • There is exactly one c with r(c) = 0
   • The number c equals 2
   Answer

   There is at least one c in (−1, 6) with r(c) = 0

2. The curve also reaches the height k = 1 three times. Why can the theorem not support the claim exactly one c?

   • The theorem promises existence only, and the picture shows three crossings
   • The number 1 is not between the end heights
   • The function r is not continuous
   Answer

   The theorem promises existence only, and the picture shows three crossings

3. A function \(f\) is continuous on \([0, 4]\), with \(f(0) = -2\) and \(f(4) = 10\). Which conclusion about \(f(x) = 3\) is justified?

   • There is at least one c in (0, 4) with f(c) = 3
   • There is exactly one c in (0, 4) with f(c) = 3
   • The number c equals 5/3
   Answer

   There is at least one c in (0, 4) with f(c) = 3

Summary

You state the conclusion as there is at least one c in the interval with f(c) = k, and you stop there.

This works because the theorem gives existence only, not the position of c or the number of solutions.

Add a second fact, such as f increasing on the interval, when you need exactly one.

This is one of four skills in topic 1.16 of Which & Why: AP Calculus.