A fixed point of a function f is a value c for which f(c) = c.
Consider h(x) = x³ - 1, where x is in the interval [1,2].
If c is a fixed point, we must have h(c) = c, so c³ - 1 = c. Rearranging this, have c³ - c - 1 = 0. So let's say g(x) = x³ - x - 1; if we can guarantee that g has exactly one zero in [1,2], then h(x) has exactly one fixed point on [1,2].
First, notice that g(1) = -1 and g(2) = 5. Since g is continuous on [1,2], it must have at least one zero on that interval by the Intermediate Value Theorem, and so h must have at least one fixed point on that interval.
But is that fixed point unique? Let's look at g'(x), which is 3x² - 1. Notice that g'(1) = 2, and g' is increasing for x > 0, so g'(x) is positive for all x in the interval [1,2]. This means that g is strictly increasing on the interval [1,2]. That's enough to guarantee that the zero of g, and therefore the fixed point of h, is unique.
( As a quick sanity check, WolframAlpha shows that the only real fixed point of h is about 1.3247. http://www.wolframalpha.com/input/?i=x%5E3+-+1+%3D+x