Don't assume that a single positive first derivative implies locally invertible; it does not. You need positive derivatives throughout an interval. We haven't introduced the trig functions yet, but you probably know what they are, so consider:
f(x) = x + x2×sin(1/x)
We know that f is bounded between x+x2 and x-x2. Both bounds have a derivative of 1 at the origin, hence f has a derivative of 1 at the origin. Let's consider its derivative near the origin.
1 + 2x×sin(1/x) - cos(1/x)
This derivative is both positive and negative arbitrarily close to the origin. There is no "stricly increasing" neighborhood about the origin; there is no interval around 0 where f can be inverted.
If f′ is nonzero and continuous at the origin, there is a small interval containing the origin, such that f′ has the same sign as f′(0). Now f is strictly increasing or decreasing on this interval, and is locally invertible. Thus f is locally invertible at any point where f′ is nonzero, provided f′ is continuous.