A continuous function cannot skip a height between its endpoint values.
The idea
A continuous curve can rise, fall, or turn back, but it cannot skip a height between its endpoint values. Every intermediate output must occur somewhere in between.
Theorem
Intermediate Value Theorem
If \(f\) is continuous on \([a,b]\) and \(k\) lies strictly between \(f(a)\) and \(f(b)\), then there is at least one \(c\in(a,b)\) such that
\[f(c)=k.\]
The target \(k\) is a height; the theorem guarantees an input \(c\) that reaches it. It does not tell us the exact input or promise that there is only one.
Drag the amber line up or down, including beyond the endpoint heights. Every height between \(f(a)\) and \(f(b)\) meets the curve; each marked input \(c\) satisfies \(f(c)=k\). Outside those heights, the theorem makes no promise.
Proving a solution exists
To show that \(f(x)=k\) has a solution, choose an interval, check continuity throughout it, and show that the endpoint outputs lie on opposite sides of \(k\). You can prove existence even when solving the equation exactly is difficult.
Example 1
A root without solving the equation
Show that \(x^3+x-1=0\) has a solution between \(0\) and \(1\).
Set \(p(x)=x^3+x-1\). This polynomial is continuous on \([0,1]\), and
\[p(0)=-1<0<1=p(1).\]
By the Intermediate Value Theorem, some \(c\in(0,1)\) satisfies \(p(c)=0\). This proves a root exists; it does not calculate it.
Example 2
An intermediate temperature
A sensor records a temperature \(T(t)\) that varies continuously, with \(T(0)=20\) and \(T(10)=80\), measured in degrees Celsius and minutes.
Since \(20<50<80\), the theorem guarantees a time \(c\in(0,10)\) when
\[T(c)=50.\]
We cannot conclude \(c=5\): continuous change need not be linear. The temperature may also reach that value more than once.
Example 3
A sign change across a break
For \(f(x)=1/x\), the endpoint values satisfy
\[f(-1)=-1<0<1=f(1).\]
But \(f\) is undefined at \(0\), so it is not continuous on \([-1,1]\). The theorem does not apply: \(x=0\) is a vertical asymptote, not a zero. Indeed, \(1/x=0\) has no solution.
