overall accuracy: 87% accuracy last 50: 86% record: 40 score: 40 selected values of the function f are shown…

overall accuracy: 87% accuracy last 50: 86% record: 40 score: 40 selected values of the function f are shown below. which statements must be true? i. there exists a c, for 0 < x < 4, such that f(c) = 9. ii. there exists a c, for 0 < x < 4, such that f(c) = 2. x 0 4 f(x) 2 10 i and ii ii only neither i only high score board: overall refresh you must have at least one to be on the board. # name record

overall accuracy: 87% accuracy last 50: 86% record: 40 score: 40 selected values of the function f are shown below. which statements must be true? i. there exists a c, for 0 < x < 4, such that f(c) = 9. ii. there exists a c, for 0 < x < 4, such that f(c) = 2. x 0 4 f(x) 2 10 i and ii ii only neither i only high score board: overall refresh you must have at least one 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 = 0), (b = 4), (f(0)=2), (f(4)=10), and (k = 9). Since (2<9<10) and assuming (f(x)) is continuous on ([0,4]), there exists a (c\in(0,4)) such that (f(c)=9).

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 = 4), (f(0)=2), (f(4)=10), so (\frac{f(4)-f(0)}{4 - 0}=\frac{10 - 2}{4}=2). So, if (f(x)) is continuous on ([0,4]) and differentiable on ((0,4)), there exists a (c\in(0,4)) such that (f^{\prime}(c)=2).

Answer:

I and II