Hi:

Assuming that you mean to say f(x) is increasing everywhere between points (0, 2) and (6, 5), we would say

The function, f(x) is increasing on the interval [0, 6], with range [2, 5]. The so called, interval, refers to the function's domain. This being the case, we can be certain that f^-1 indeed exists. The domain of f^-1 is identical to the range of f and, conversely, the range of f^-1 is equal to the domain of f. Hence,

f^-1 is increasing on the interval [2, 5] (closed interval by the way), with range [0, 6].

As a note on 'end points', we were given that the graph of f(x) has endpoints (0,2) and (6,5). It therefore stands to reason that f^-1(x) extends between endpoints (2, 0) and (5, 6). For every point (x, y) satisfying y=f(x), (y,x) will satisfy f^-1. Because f maps 0 into 2 (i.e., f(0)=2), f^-1 must map 2 back into 0 (that is what an inverse does -- it reverses each mapping under the 'original' function, f. And, vice versa). Likewise, f(6)=5 implies that f^-1(5) = 6.

Regards,

Rich B.