Let f:[-2,2] rightarrow mathbb{R} be a contin | vedclass

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(D) continuous function $f: I \rightarrow \mathbb{R}$ defined on an interval $I$ that takes only irrational values must be a constant function. If $f(

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$f(\sqrt{2}-1)=\sqrt{2}$

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(D) continuous function $f: I \rightarrow \mathbb{R}$ defined on an interval $I$ that takes only irrational values must be a constant function. If $f(x)$ were not constant, by the Intermediate Value Theorem, it would take all values between any two values it assumes. Since the set of rational numbers is dense in $\mathbb{R}$, any non-constant continuous function must take rational values. Given $f(\sqrt{2})=\sqrt{2}$, and since $f(x)$ is constant, we have $f(x)=\sqrt{2}$ for all $x \in [-2,2]$. Therefore, $f(\sqrt{2}-1)=\sqrt{2}$.

Let $f:[-2,2] \rightarrow \mathbb{R}$ be a continuous function such that $f(x)$ assumes only irrational values. If $f(\sqrt{2})=\sqrt{2},$ then

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