Suppose [math]. In this problem, we will show that [math] has exactly one root (or zero) in the interval [math].

(a) First, we show that [math] has a root in the interval [math]. Since [math] is a function on the interval [math] and [math] and [math] , the graph of [math] must cross the [math]-axis at some point in the interval [math] by the . Thus, [math] has at least one root in the interval [math].

(b) Second, we show that [math] cannot have more than one root in the interval [math] by a thought experiment. Suppose that there were two roots [math] and [math] in the interval [math] with [math]. Then [math] . Since [math] is on the interval [math] and on the interval [math], by there would exist a point [math] in interval [math] so that [math]. However, the only solution to [math] is [math] , which is not in the interval [math], since [math]. Thus, [math] cannot have more than one root in [math].

(Note: where the problem asks you to make a choice select the weakest choice that works in the given context. For example "continuous" is a weaker condition than "polynomial" because every polynomial is continuous but not vice-versa. Rolle's theorem is a weaker theorem than the mean value theorem because Rolle's theorem applies to fewer cases.)