Question

1. on which of the given intervals is the average rate of change of the given graph equal to 0?

a. $-2,-1$
b. $-3,0$
c. $0,2$
d. $-1,2$
e. $-2,2$

1. what are the domain and range of the given

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is given by \(\frac{f(b)-f(a)}{b - a}\). If the average rate of change is \(0\), then \(f(b)-f(a)=0\), i.e., \(f(a)=f(b)\).

Step2: Evaluate each interval

• For interval \([-2,-1]\):

Let's assume the function values. If \(x=-2\) and \(x = - 1\), from the graph (by visual inspection of the \(y\) - values corresponding to \(x=-2\) and \(x=-1\)), \(f(-2)
eq f(-1)\).

• For interval \([-3,0]\):

By visual inspection of the \(y\) - values corresponding to \(x=-3\) and \(x = 0\) on the graph, \(f(-3)
eq f(0)\).

• For interval \([0,2]\):

By visual inspection of the \(y\) - values corresponding to \(x = 0\) and \(x=2\) on the graph, \(f(0)
eq f(2)\).

• For interval \([-1,2]\):

By visual inspection of the \(y\) - values corresponding to \(x=-1\) and \(x = 2\) on the graph, \(f(-1)
eq f(2)\).

• For interval \([-2,2]\):

Let \(a=-2\) and \(b = 2\). By visual inspection of the \(y\) - values (since the graph is symmetric about the \(y\) - axis (the vertex is on the \(y\) - axis)), \(f(-2)=f(2)\). Then \(\frac{f(2)-f(-2)}{2-(-2)}=\frac{0}{4}=0\).

Answer:

e. \([-2,2]\)