Unit 1: Limits and Their Properties

contintinuity

1. f(a) exists 2. lim_(x→a) f(x) exists 3. lim_(x→a) f(x) = f(a)

IVT phrasing - not necessarily continuous

The function f(x) has a domain of [−9,9] and is plotted below such that the portion of the graph on the interval (−4,2) is hidden from view. Given that f(-4)=-3 and f(2)=-7, determine what conclusions can be drawn based on the Intermediate Value Theorem on the interval (-4, 2). Since the function f(x) is not necessarily continuous on its domain, the Intermediate Value Theorem cannot be applied. Therefore, there may not exist a value of c in the interval (-4, 2) where f(c)=-4.

IVT phrasing - continuous, no c value in interval

The function f(x) is continuous on its domain of [−9,9] and is plotted below such that the portion of the graph on the interval (-1, 6) is hidden from view. Given that f(-1)=4 and f(6)=2, determine what conclusions can be drawn based on the Intermediate Value Theorem on the interval (-1, 6). Since the function f(x) is continuous on its domain, the Intermediate Value Theorem can be applied and states that for any value kk in the interval (2,4), there exists a value c, where -1 < c < 6, such that f(c)=k. Therefore, there may not exist a value of cc in the interval (-1, 6) where f(c)=0.

IVT phrasing - continuous, c exists in interval

The function f(x) is continuous on its domain of [−9,9] and is plotted below such that the portion of the graph on the interval (−1,4) is hidden from view. Given that f(-1)=-3 and f(4)=2, determine what conclusions can be drawn based on the Intermediate Value Theorem on the interval (-1, 4). Since the function f(x) is continuous on its domain, the Intermediate Value Theorem can be applied and states that for any value k in the interval (-3, 2), there exists a value c, where -1 < c < 4, such that f(c)=k. Therefore, there exists a value of c in the interval (−1,4) where f(c)=1.