You Can't Climb Without Crossing Every Height

This scene checks the intermediate value theorem. Move the target height and find where the graph crosses it.

An unbroken path cannot skip the middle. Going from low to high, it must pass every height between. If the graph breaks, the promise fails.

For example, watch the column of a 🌡️ thermometer. Rising from 10° in the morning to 20° at noon, it must pass 15°. The column never teleports.

Examples & Exercises

Use the Intermediate Value Theorem to show that $x^3 + x - 1 = 0$ has a real root in $(0,1)$ .

f(0) = -1 < 0 < 1 = f(1) \ \Longrightarrow\ \exists\, \xi \in (0,1):\ f(\xi) = 0

c = 0

— the point

\xi \approx 0.6823

where the curve steps on height

0

Show proof

$f(x) = x^3 + x - 1$ is a polynomial, hence continuous on $[0,1]$ .
$f(0) = -1 < 0$ and $f(1) = 1 > 0$ .
By the Intermediate Value Theorem there is $\xi \in (0,1)$ with $f(\xi) = 0$ (strictly inside, since $f(0), f(1) \neq 0$ ). Numerically $\xi \approx 0.6823278038$ .

Show that $\sin x = \frac12$ has a solution in $(0, \frac{\pi}{2})$ .

\sin 0 = 0 < \tfrac12 < 1 = \sin \tfrac{\pi}{2} \ \Longrightarrow\ \exists\, \xi:\ \sin \xi = \tfrac12

c = 0.5

— the crossing is at

\xi = \pi/6 \approx 0.5236

Show proof

$\sin x$ is continuous on $[0, \frac{\pi}{2}]$ .
$\sin 0 = 0 < \frac12$ and $\sin \frac{\pi}{2} = 1 > \frac12$ .
By the Intermediate Value Theorem there is $\xi \in (0, \frac{\pi}{2})$ with $\sin \xi = \frac12$ .
The witness is unique: for $0 \le x < y \le \frac{\pi}{2}$ , $\sin y - \sin x = 2\cos\frac{x+y}{2}\sin\frac{y-x}{2} > 0$ , since $\frac{x+y}{2} \in [0, \frac{\pi}{2})$ gives $\cos\frac{x+y}{2} > 0$ and $\frac{y-x}{2} \in (0, \frac{\pi}{4}]$ gives $\sin\frac{y-x}{2} > 0$ — so $\sin$ is strictly increasing on $[0, \frac{\pi}{2}]$ .
Since $\sin \frac{\pi}{6} = \frac12$ , uniqueness forces $\xi = \frac{\pi}{6}$ .

Show directly that $f(x) = \begin{cases} 0 & x < \frac12 \\ 1 & x \geq \frac12 \end{cases}$ on $[0,1]$ never takes the value $c = \frac12$ , and name the IVT hypothesis that fails.

f([0,1]) = \{0, 1\} \not\ni \tfrac12

The continuous case (

c = 0

) — drop the hypothesis and the crossing may vanish.

Show proof

For $x < \frac12$ , $f(x) = 0 \neq \frac12$ ; for $x \geq \frac12$ , $f(x) = 1 \neq \frac12$ . So no $x$ satisfies $f(x) = \frac12$ , even though $f(0) = 0 < \frac12 < 1 = f(1)$ .
The broken hypothesis is continuity at $x = \frac12$ : the left-hand limit is $0$ while $f(\frac12) = 1$ . This counterexample shows the IVT conclusion need not hold without its hypothesis.