Let f(t) = 1/t. Find a value of t such that the average rate of change of f(t) from 1 to t equals -1/13.

Solution:
1. The average rate of change of a function f(t) from t = a to t = b is given by:
\frac{f(b) - f(a)}{b - a}

2. Given:
* Function: f(t) = \frac{1}{t}
* Interval: [1, t]
* Average rate of change: -\frac{1}{13}

3. Substitute the given values into the formula:
\frac{f(t) - f(1)}{t - 1} = -\frac{1}{13}

4. Evaluate f(1):
f(1) = \frac{1}{1} = 1

5. Substitute f(1) and f(t):
\frac{\frac{1}{t} - 1}{t - 1} = -\frac{1}{13}

6. Simplify the numerator:
\frac{1 - t}{t(t - 1)} = -\frac{1}{13}

7. Multiply both sides by t(t - 1):
1 - t = -\frac{t(t - 1)}{13}

8. Distribute and simplify:
1 - t = -\frac{t^2 - t}{13}
13(1 - t) = -t^2 + t
13 - 13t = -t^2 + t

9. Rearrange into a standard quadratic equation:
t^2 - 14t + 13 = 0

10. Factor the quadratic equation:
(t - 13)(t - 1) = 0

11. Solve for t:
t = 13 \quad \text{or} \quad t = 1

12. Since we are looking for the interval from 1 to t and t = 1 does not change the interval, we discard t = 1.

Therefore, the solution is t = 13.