Solution
= Solution
\b[True.] Part (a) shows that $f$ is strictly monotone. Suppose it is increasing and let $y_0=f(x_0)$. Given $\epsilon>0$,
$$
f(x_0-\epsilon)<y_0<f(x_0+\epsilon).
$$
Choose $\delta>0$ smaller than both distances from $y_0$ to the two outer values. If $|y-y_0|<\delta$, monotonicity forces
$$
x_0-\epsilon<f^{-1}(y)<x_0+\epsilon.
$$
\b[Thus $f^{-1}$ is continuous at $y_0$. The decreasing case is identical with inequalities reversed.]