selected values of the continuous function f are shown below. which statements must be true? o i. there…

selected values of the continuous function f are shown below. which statements must be true? o i. there exists a c, for 0 < x < 2, such that f(c) = 3. o ii. there exists a c, for 0 < x < 2, such that f(c) = 3. x 0 2 f(x) 5 11 ii only i and ii neither i only

selected values of the continuous function f are shown below. which statements must be true? o i. there exists a c, for 0 < x < 2, such that f(c) = 3. o ii. there exists a c, for 0 < x < 2, such that f(c) = 3. x 0 2 f(x) 5 11 ii only i and ii neither i only

Answer

Explanation:

Step1: Apply Intermediate - Value Theorem

The Intermediate - Value Theorem states that if (y = f(x)) is continuous on the closed interval ([a,b]) and (k) is a number between (f(a)) and (f(b)), then there exists at least one number (c) in the open interval ((a,b)) such that (f(c)=k). Here, (a = 0), (b = 2), (f(0)=5), (f(2)=11), and (k = 3). Since (3) is not between (5) and (11), the Intermediate - Value Theorem does not guarantee the existence of a (c\in(0,2)) such that (f(c)=3).

Step2: Apply Mean - Value Theorem

The Mean - Value Theorem states that if (y = f(x)) is continuous on the closed interval ([a,b]) and differentiable on the open interval ((a,b)), then there exists at least one number (c\in(a,b)) such that (f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}). Here, (a = 0), (b = 2), (f(0)=5), (f(2)=11), so (\frac{f(2)-f(0)}{2 - 0}=\frac{11 - 5}{2}=\frac{6}{2}=3). Since (f(x)) is continuous on ([0,2]) and assuming it is differentiable on ((0,2)) (a necessary condition for the Mean - Value Theorem, and since no information to the contrary is given, we can apply it), there exists a (c\in(0,2)) such that (f^{\prime}(c)=3).

Answer:

II only