This implies that $f$ is strictly increasing on $(a, b)$. Therefore, for all $x_1 > x_0$ in $(a, b)$, we have $f(x_1) > f(x_0)$. This contradicts the fact that $f(a) = f(b) = 0$, because $f(b)$ cannot be equal to $f(a)$ if $f$ is strictly increasing. Therefore,
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