overall accuracy: 88.6% record: 34 score: 34 selected values of the continuous function f are shown below…

overall accuracy: 88.6% record: 34 score: 34 selected values of the continuous function f are shown below. which statements must be true? o i. there exists a c, for 2 < x < 5, such that f(c) = 4. o ii. there exists a c, for 2 < x < 5, such that f(c) = 2. x 2 5 f(x) 1 7 i only i and ii neither ii only high score board: overall refresh you must have at least 100 to be on the board. # name record

overall accuracy: 88.6% record: 34 score: 34 selected values of the continuous function f are shown below. which statements must be true? o i. there exists a c, for 2 < x < 5, such that f(c) = 4. o ii. there exists a c, for 2 < x < 5, such that f(c) = 2. x 2 5 f(x) 1 7 i only i and ii neither ii only high score board: overall refresh you must have at least 100 to be on the board. # name record

Answer

Explanation:

Step1: Apply Intermediate - Value Theorem

The Intermediate - Value Theorem states that if a function (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 = 2), (b = 5), (f(2)=1), (f(5)=7), and (k = 4). Since (1<4<7) and (f(x)) is continuous, there exists a (c\in(2,5)) such that (f(c)=4).

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 = 2), (b = 5), (f(2)=1), (f(5)=7), so (\frac{f(5)-f(2)}{5 - 2}=\frac{7 - 1}{3}=2). Since (f(x)) is continuous on ([2,5]) and differentiable (because we can talk about (f^{\prime}(c))), by the Mean - Value Theorem, there exists a (c\in(2,5)) such that (f^{\prime}(c)=2).

Answer:

I and II