Check the Intermediate Value Theorem conditions
An unbroken curve with two known ends
The solid curve is the graph of \(f(x) = x^3 + 2x - 5\) on the closed interval from \(x = 1\) to \(x = 2\), which includes both ends.
Drag the orange point from one end to the other. The pen never leaves the paper, because the curve has no gap and no jump.
The orange dots mark the two ends, at height \(-2\) when \(x = 1\) and height 7 when \(x = 2\). Because the curve is unbroken, it passes through every height between \(-2\) and 7, and the Intermediate Value Theorem guarantees this.
Before you use the theorem, you must confirm two facts. The curve is unbroken on the whole interval, and you know the heights at the two ends.
Any function on a closed interval
The curve is any function \(f\) on the closed interval from \(a\) to \(b\). Orange dots mark its ends, at the heights \(\cy{f(a)}\) and \(\cy{f(b)}\).
Drag the slider for the right end \(b\). The right dot moves along the curve, so its height \(\cy{f(b)}\) changes.
The curve stays in one piece from \(a\) to \(b\), so \(f\) is continuous at every point of \([a, b]\).
Notation
Name the interval. The closed interval from \(a\) to \(b\) contains \(a\), \(b\) and every number between them.
\[[a, b] = \{x : a \le x \le b\}\]Check that f is continuous at every point of the interval. A polynomial is continuous everywhere. A fraction is continuous where its bottom is not 0. A square root is continuous where the number inside it is at least 0.
At an end point, continuity uses only the side inside the interval, because no points of the interval lie on the other side.
\[\begin{aligned} \lim_{x \to a^+} f(x) &= f(a) \\ \lim_{x \to b^-} f(x) &= f(b) \end{aligned}\]Compute the heights at the two ends, because the theorem guarantees only the heights between f(a) and f(b).
\[\cy{f(a)} \quad\text{and}\quad \cy{f(b)}\]
When it fails
Press Put a break at c. The curve now jumps at a point \(c\) between \(a\) and \(b\), so the pen must lift there. The red segment shows the jump, which lies inside the interval while \(b\) is to the right of \(c\).
Drag the right end \(b\) to the left of \(c\). The interval no longer holds the break, so \(f\) is continuous on it and the conditions hold again.
Example 1: a square root
Check the conditions of the theorem for \(f(x) = \sqrt{x + 3} + x^2\) on \([-2, 1]\), and compute the end heights.
The root is defined only when the number inside it is at least 0, which means \(x \ge -3\). The interval \([-2, 1]\) lies to the right of the faint line at \(x = -3\), so the root is defined on all of it.
\[x + 3 \ge 0 \iff x \ge -3\]A sum of continuous functions is continuous. The root and \(x^2\) are continuous on \([-2, 1]\), so \(f\) is continuous on \([-2, 1]\) and the pen stays on the paper.
Compute the end heights.
\[\begin{aligned} f(-2) &= \sqrt{1} + 4 = \cy{5} \\ f(1) &= \sqrt{4} + 1 = \cy{3} \end{aligned}\]
Example 2: a piecewise function
Check the conditions for \(f(x) = \begin{cases} 2x - 1 & x < 2 \\ x^2 + 1 & x \ge 2 \end{cases}\) on \([0, 3]\).
The formula changes at x = 2. Each piece is a polynomial, so each piece is continuous on its own.
A tempting wrong move: Both pieces are continuous, so \(f\) is continuous on \([0, 3]\).
\[\cr{f \text{ is continuous on } [0, 3]}\]The left piece approaches height 3 at x = 2, but the right piece starts at height 5. The red segment is a jump of 2, so the pen lifts.
Continuity at x = 2 needs the left limit, the right limit and f(2) to agree. The two limits differ, so f is not continuous at 2 and the conditions fail on [0, 3].
\[\begin{aligned} \lim_{x \to 2^-} (2x - 1) &= \cy{3} \\ \lim_{x \to 2^+} (x^2 + 1) &= \cy{5} \end{aligned}\]
Practice
Check the conditions for \(f(x) = \dfrac{3x}{x - 4}\) on \([5, 8]\), and compute the end heights.
Look for a break in the picture. Is the break inside the interval or outside it?
Solution
A fraction is continuous where its bottom is not 0. The bottom is 0 only at the dashed line \(x = 4\), outside \([5, 8]\), so \(f\) is continuous on \([5, 8]\).
Compute the end heights.
\[\begin{aligned} f(5) &= \frac{15}{1} = \cy{15} \\ f(8) &= \frac{24}{4} = \cy{6} \end{aligned}\]
Questions
The graph shows f on [0, 4]. A filled dot is a value of f. An open ring is a height that the curve approaches but does not take.
Why can the Intermediate Value Theorem not be used on [0, 4]?
- The graph has a jump at x = 2, inside the interval
- The heights at the two ends are both positive
- The function is not defined at x = 2
Answer
The graph has a jump at x = 2, inside the interval
On which closed interval is this graph continuous at every point?
- [2, 4]
- [0, 2]
- [0, 4]
Answer
[2, 4]
On which interval is \(f(x) = \dfrac{\sqrt{x - 2}}{x - 5}\) continuous at every point?
- [2, 4]
- [2, 5]
- [3, 6]
- [0, 4]
Answer
[2, 4]
Summary
You check that the function is continuous on the closed interval, and you compute the two end heights.
The Intermediate Value Theorem needs both facts, because only an unbroken curve must pass through every height between its end heights.
Do these two steps before you apply the theorem to an equation.
This is one of four skills in topic 1.16 of Which & Why: AP Calculus.