there exists a value c in the open interval (-2,0) such that f(c)=1.\nanswer\nthe statement must be…

there exists a value c in the open interval (-2,0) such that f(c)=1.\nanswer\nthe statement must be true\nthe statement must be false\nthe statement could be either true or false

there exists a value c in the open interval (-2,0) such that f(c)=1.\nanswer\nthe statement must be true\nthe statement must be false\nthe statement could be either true or false

Answer

Explanation:

Step1: Calculate the average rate of change

The average rate of change of a function (y = f(x)) over the interval ([a,b]) is given by (\frac{f(b)-f(a)}{b - a}). Here, (a=-2), (b = 0), (f(-2)=3) and (f(0)=1). So, (\frac{f(0)-f(-2)}{0-(-2)}=\frac{1 - 3}{2}=\frac{-2}{2}=-1).

Step2: Analyze the relationship between average rate of change and derivative

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}). But we only know the values of the function at (x=-2), (x=-1) and (x = 0). We have no information about the continuity and differentiability of (f(x)) on ((-2,0)). Just because the average rate of change over ([-2,0]) is (- 1), we cannot be sure that there is a (c\in(-2,0)) such that (f^{\prime}(c)=1). For example, if (f(x)) is a non - differentiable function (e.g., a piece - wise function with a sharp corner) or a function whose derivative never equals (1) in the interval ((-2,0)) (even if it is differentiable), the statement (f^{\prime}(c)=1) for some (c\in(-2,0)) may not hold.

Answer:

the statement could be either true or false