Exercise 110Calculus › Single-variable calculus › Continuity › Intermediate Value TheoremHigh schoolAdded 8 Nov 2026

Continuous functions with f^2=1 on an interval

#IVT#sign#constant functions

Statement

Let $I$ be an interval of $\mathbb{R}$ and $f:I\to\mathbb{R}$ continuous with $(f(x))^2=1$ for all $x\in I$ . Show that $f=1$ or $f=-1$ .

Hint

Assume $f$ takes both values and apply the IVT to reach $f(z)=0$ , contradicting $(f(z))^2=1$ .

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Solution

At each $x\in I$ , $(f(x))^2=1$ gives $f(x)\in\{-1,1\}$ . Suppose $f$ is neither identically $1$ nor identically $-1$ . Then there exist $x_0,y_0\in I$ with $f(x_0)=1$ and $f(y_0)=-1$ .

Since $I$ is an interval, it contains the whole segment between $x_0$ and $y_0$ . The IVT applied to the continuous $f$ on this segment provides $z$ between $x_0$ and $y_0$ with $f(z)=0$ . But then $(f(z))^2=0\neq1$ : contradiction.

Conclusion. $f$ is constant on $I$ , with value $1$ or $-1$ .

Continuous functions with f^2=1 on an interval

Statement

Solution

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