Recognize when the theorem cannot decide
Two curves with a jump
Two curves on the interval from 0 to 4 both start at height 1, end at height 7, and jump at x = 2. The orange line is at height 4, which is between 1 and 7.
On curve A, the jump goes from height 3 to height 5, and the red segment marks the heights that it skips. The line at 4 passes through this gap, so curve A never takes the value 4.
Press Curve B. The left piece rises through the height 4 at x = 1.5, so curve B takes the value 4.
The end heights and the jump are the same for both curves, so they alone do not decide whether a curve takes 4.
A failed condition gives no answer
The theorem has two conditions: \(f\) is continuous on \([a, b]\), and \(k\) is strictly between \(f(a)\) and \(f(b)\). When both hold, the curve takes the value \(\cy{k}\).
Press Put a jump in the curve. The first condition fails, and the red segment marks the heights that the curve skips. Press Move k above both ends. The second condition fails.
Both failed cases here miss the line, but curve B in the opening had a jump and still took the value 4. A failed condition means only that the theorem gives no answer.
Notation
If both conditions hold, the theorem gives its conclusion: some number \(c\) in \((a, b)\) has \(f(c) = k\).
\[\begin{aligned} &\text{continuous on } [a, b] \\ &k \text{ strictly between } f(a), f(b) \\ &\Rightarrow \exists\, c \in (a, b) : f(c) = k \end{aligned}\]If a condition fails, the theorem does not apply. Name the condition that fails. Either f has a break in [a, b], or k is not strictly between f(a) and f(b).
When the theorem does not apply, the equation f(x) = k may or may not have a solution. Decide with another method, such as algebra. Here the jump stays, but a lower k meets the left piece.
\[\begin{aligned} \text{a condition fails} &\Rightarrow \\ \text{no conclusion} & \end{aligned}\]
When it fails
With no jump, the curve is continuous and the theorem promises the value \(k\). Drag the jump size up. The right piece moves up, and a red segment marks the gap.
After a small jump the curve can still cross the line, but the theorem no longer promises it. When the gap contains the height k, the curve misses the value.
Example 1: a break inside the interval
Can the theorem show that \(f(x) = \dfrac{3}{x - 1}\) takes the value 0 on \([0, 4]\)?
Compute the end heights. The number 0 is strictly between them, so the second condition holds.
\[f(0) = \cy{-3} < 0 < \cy{1} = f(4)\]The denominator is 0 at x = 1, which is inside [0, 4], so the curve breaks at the dashed line. The function is not continuous on [0, 4], so the first condition fails.
The theorem gives no answer, so use algebra. A fraction with numerator 3 is never 0, so the equation has no solution.
\[\frac{3}{x - 1} = 0 \text{ has no solution}\]
Example 2: a jump that still takes the value
Can the theorem show that \(h(x) = \begin{cases} x + 1 & 0 \le x < 2 \\ 8 - x & 2 \le x \le 4 \end{cases}\) takes the value 2?
Compute the end heights. The number 2 is strictly between them, so the second condition holds.
\[h(0) = \cy{1} < 2 < \cy{4} = h(4)\]The formula changes at \(x = 2\). The left piece approaches 3, but \(h(2) = 6\), so \(h\) has a jump there and the first condition fails.
\[\lim_{x \to 2^-} h(x) = 3 \ne 6 = h(2)\]A tempting wrong move: The theorem does not apply, so h(x) = 2 has no solution.
\[\cr{\text{no solution}}\]The left piece rises through the height 2 at x = 1, so h takes the value 2 even though the theorem gives no answer.
Solve with the formula. The equation \(x + 1 = 2\) gives \(x = 1\), which is in \([0, 2)\).
\[h(1) = 1 + 1 = \cy{2}\]
Practice
Can the theorem show that \(f(x) = x^2 - 4x\) takes the value \(-4\) on \([0, 3]\)? Does \(f\) take it?
Compare −4 with the end heights. Then find the lowest point in the picture.
Solution
The function \(f\) is continuous, because it is a polynomial. The number \(-4\) is below both end heights, so the second condition fails.
\[\begin{aligned} f(0) &= 0 \\ f(3) &= -3 \\ -4 &< -3 \end{aligned}\]The theorem gives no answer, so use algebra. Since \(x^2 - 4x = (x - 2)^2 - 4\), the lowest point is at \(x = 2\), where \(f\) takes the value \(-4\).
\[f(2) = 4 - 8 = \cy{-4}\]
Questions
The graph shows m on [0, 5]. An open ring is a height that the curve approaches but does not take. A filled dot is a value of m.
Which height between the end heights does the curve miss?
- 3
- 1
- 5
Answer
3
The height 1 is between the end heights 0 and 7, and the curve takes it. What does this show?
- A curve with a jump can take a value even though the theorem does not apply
- The theorem applies, because the curve takes the value 1
- Every height between 0 and 7 is taken
Answer
A curve with a jump can take a value even though the theorem does not apply
Let \(g(x) = \dfrac{x^2 - 1}{x - 1}\) for \(x \ne 1\), and let \(g(1) = 3\). What is true for the value 2 on \([0, 2]\)?
- The theorem does not apply, and g takes no value 2
- The theorem gives a number c in (0, 2) with g(c) = 2
- The theorem does not apply, but c = 1 solves the equation
Answer
The theorem does not apply, and g takes no value 2
Summary
If a condition of the theorem fails, you name it and say that the theorem gives no answer.
A failed condition removes the promise of the theorem, but a curve with a jump can still take the value.
To decide whether f takes the value k, you solve f(x) = k or read the graph.
This is one of four skills in topic 1.16 of Which & Why: AP Calculus.