Question

t g(t) -5 6 -4 4 -3 2 -2 1 -1 -1 0 -2 1 -3 2 1 3 3 4 6 5 9 over which interval does g have an average rate of change of zero? choose 1 answer: a -2 ≤ t ≤ 2 b 0 ≤ t ≤ 4 c 1 ≤ t ≤ 3 d -5 ≤ t ≤ 1

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \(y = g(t)\) over the interval \([a,b]\) is given by \(\frac{g(b)-g(a)}{b - a}\). We want to find when \(\frac{g(b)-g(a)}{b - a}=0\), which implies \(g(b)=g(a)\).

Step2: Check each interval

• For interval \(A\): \(-2\leq t\leq2\)

Here \(a=-2\) and \(b = 2\). \(g(-2)=1\) and \(g(2)=1\). Then the average rate of change is \(\frac{g(2)-g(-2)}{2-(-2)}=\frac{1 - 1}{4}=0\).

• For interval \(B\): \(0\leq t\leq4\)

\(a = 0\), \(b=4\). \(g(0)=-2\) and \(g(4)=6\). The average rate of change is \(\frac{6-(-2)}{4-0}=\frac{8}{4}=2
eq0\).

• For interval \(C\): \(1\leq t\leq3\)

\(a = 1\), \(b = 3\). \(g(1)=-3\) and \(g(3)=3\). The average rate of change is \(\frac{3-(-3)}{3 - 1}=\frac{6}{2}=3
eq0\).

• For interval \(D\): \(-5\leq t\leq1\)

\(a=-5\), \(b = 1\). \(g(-5)=6\) and \(g(1)=-3\). The average rate of change is \(\frac{-3 - 6}{1-(-5)}=\frac{-9}{6}=-\frac{3}{2}
eq0\).

Answer:

A. \(-2\leq t\leq2\)