Question

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

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