Identify a guaranteed intermediate value
A horizontal line across an unbroken curve
The curve is the graph of \(f(x) = x^3 + x + 1\) from \(x = 0\) to \(x = 2\). It starts at height 1, ends at height 11, and has no breaks.
Drag the slider to move the horizontal line up and down. The height of the line is called \(\cy{k}\).
When \(k\) is between 1 and 11, the line turns orange and meets the curve at an orange dot. For \(k = 6\) the dot is at \(x \approx 1.52\).
When k is below 1 or above 11, the line is faint and does not meet the curve between x = 0 and x = 2.
Because the curve has no breaks, it must pass through every height between 1 and 11.
Any end heights and any target
The curve is any unbroken curve from the point \((a, f(a))\) to the point \((b, f(b))\). The orange line is at the height \(\cy{k}\).
The two faint dashed lines mark the end heights. Drag the slider. Whenever the line is between the dashed lines, it meets the curve at an orange dot.
Press Make the curve fall. The curve now falls from \(f(a)\) to \(f(b)\), and the line still meets it for every height between the dashed lines.
The curve can rise or fall, so the order of the end heights does not matter. Every height between them is taken.
Notation
The target height \(\cy{k}\) is the height that you want the function to take. It is strictly between the end heights when it is larger than one of them and smaller than the other.
\[\begin{aligned} f(a) < k &< f(b) \\ \text{or } f(b) < k &< f(a) \end{aligned}\]The Intermediate Value Theorem has two conditions: \(f\) is continuous on the closed interval \([a, b]\), and \(k\) is strictly between \(f(a)\) and \(f(b)\). Then at least one number \(c\) in the open interval \((a, b)\) has \(f(c) = k\).
\[\exists\, c \in (a, b) : f(c) = \cy{k}\]To show that an equation has a solution, move all terms to one side and call the result \(f\). The target is then \(k = 0\), so the end heights must be on opposite sides of 0.
\[\begin{aligned} g(x) = h(x) &\iff \\ f(x) = g(x) - h(x) &= 0 \end{aligned}\]
When it fails
Press Raise k above both ends. The line is now above both dashed lines, so \(k\) is larger than both end heights.
The red label says that k is not between them, and the line misses the curve between a and b. The theorem then makes no promise, so check this condition first.
Example 1: a height inside the range
Does \(f(x) = x^2 + 2x\) take the value 10 on \([1, 4]\)?
The function is a polynomial, so it is continuous on [1, 4]. Compute the end heights.
\[f(1) = \cy{3} \qquad f(4) = \cy{24}\]The target 10 is larger than 3 and smaller than 24, so it is strictly between the end heights.
\[3 < \cy{10} < 24\]Both conditions hold, so the theorem gives at least one \(c\) in \((1, 4)\) with \(f(c) = 10\). The orange dot marks it.
\[\exists\, c \in (1, 4) : c^2 + 2c = 10\]
Example 2: a falling curve
Does \(f(x) = 12 - x^2\) take the value 7 on \([0, 3]\)?
The function is a polynomial, so it is continuous on [0, 3]. Compute the end heights.
\[f(0) = \cy{12} \qquad f(3) = \cy{3}\]A tempting wrong move: The statement 12 < 7 < 3 is false, so 7 is not between the end heights.
\[\cr{12 < 7 < 3}\]The curve falls from 12 to 3, and the line at 7 crosses it. Between does not require the first end height to be the smaller one.
The target 7 is larger than 3 and smaller than 12, so it is strictly between the end heights. The theorem gives at least one \(c\) in \((0, 3)\) with \(f(c) = 7\).
\[3 < \cy{7} < 12\]Here \(12 - c^2 = 7\) gives \(c = \sqrt{5} \approx 2.24\).
Practice
Show that the equation \(2^x = 3x\) has a solution in \([1, 4]\).
Move all terms to one side, so that the target is 0. Drag the pen point to each end. Is the curve below the x-axis at one end and above it at the other?
Solution
Define \(f(x) = 2^x - 3x\). It is continuous because it is a difference of continuous functions. A solution of the equation is a number \(c\) with \(f(c) = 0\).
\[f(x) = 2^x - 3x\]Compute the end heights. One is below 0 and the other is above 0, so 0 is strictly between them.
\[f(1) = \cy{-1} \qquad f(4) = \cy{4}\]Because \(-1 < 0 < 4\), the theorem gives at least one \(c\) in \((1, 4)\) with \(f(c) = 0\). This number solves \(2^x = 3x\).
Questions
The graph shows \(s(x) = (x - 2)^2 + 1\) on \([0, 5]\). The orange dots mark the ends.
Which value does the theorem guarantee that s takes on [0, 5]?
- 7
- 12
- 0
- 1
Answer
7
The orange line is at k = 8. Which inequality shows that the theorem applies?
- 5 < 8 < 10
- 1 < 8 < 10
- 8 < 5 < 10
Answer
5 < 8 < 10
For \(f(x) = x^3 - 2x\) on \([1, 3]\), which target value does the theorem guarantee?
- 0
- −2
- 22
- 27
Answer
0
Summary
You compute the two end heights and check that the target value is strictly between them, in either order.
This works because an unbroken curve must pass through every height between its two end heights.
Use it to show that an equation f(x) = k has a solution in an interval, without solving the equation.
This is one of four skills in topic 1.16 of Which & Why: AP Calculus.