Notes on first semester calculus

Proof. First, check that f is one-to-one, so that f−1 exists. Given two points in the domain of f , x = t, we want to show f(x) = f(t). From x = t, there are two cases: x < t or x > t. If x < t, then f(x) < f(t) because f is increasing, and similarly if t < x, then f(t) < f(x). In either case, f(x) = f(t). Second, suppose, toward a contradiction, that f−1(x) is not increasing. This means the following (the “negation” of the definition of increasing): there are two points (a, b) and (c, d) on the graph of f−1 with a < c and b ≥ d. However, then there are two points (b, a) and (d, c) on the graph of f with b ≥ d and a < c. The case b = d and a < c is impossible since then f would fail the vertical line test, but the case b > d and a < c is also excluded since f is increasing. This contradiction shows f−1(x) must be increasing.