 Solution

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