TJ math tests are harder than class.Practice at test level.

Answer

Full solution

1. Compute the endpoint values, because the average rate of change only uses the two endpoints: $f(1) = 1 - 6 = -5$ and $f(5) = 25 - 30 = -5$.

2. Find the average rate of change: $\frac{f(5) - f(1)}{5 - 1} = \frac{-5 - (-5)}{4} = 0$. So the first half of the statement is correct.

3. Now test the conclusion. A constant function would give the same output at every input in the interval, so check one input in the middle: $f(3) = 9 - 18 = -9 \ne -5$.

4. Since $f(3) \ne f(1)$, $f$ is not constant. In fact $f$ falls from $-5$ to $-9$ on $[1,3]$ and climbs back to $-5$ on $[3,5]$; the drops and gains cancel, which is why the average rate is $0$.

5. Conclusion: an average rate of change of $0$ only tells you the endpoints have equal outputs. It says nothing about what happens between them, so the statement is False.

Answer

Full solution

1. Check that both endpoints are in the domain, because the formula needs actual outputs. The only excluded value is $x = 3$, and $3$ is not in $[4, 7]$, so we are safe.

2. Evaluate at the left endpoint: $f(4) = \frac{2(4)+1}{4-3} = \frac{9}{1} = 9$.

3. Evaluate at the right endpoint: $f(7) = \frac{2(7)+1}{7-3} = \frac{15}{4}$.

4. Take the change in output: $f(7) - f(4) = \frac{15}{4} - 9 = \frac{15}{4} - \frac{36}{4} = -\frac{21}{4}$.

5. Divide by the change in input, $7 - 4 = 3$: $\frac{-21/4}{3} = -\frac{21}{12} = -\frac{7}{4}$. The negative sign makes sense because $f$ drops from $9$ to $3.75$ as $x$ moves right.