We are given the function $f(t) = t - 3t$ on the interval $[-1, 5]$, and need to find the absolute maximum and absolute minimum values on this interval.

However, the expression for the function seems incomplete. Based on what is provided, if we assume that the function is $f(t) = t - 3t^2$ (a common type of quadratic form), I will proceed with that assumption and solve the problem for this function. Please correct me if the function is different.

Step 1: Find the critical points

To find the critical points, we first compute the derivative of $f(t)$.

$f'(t) = \frac{d}{dt}(t - 3t^2) = 1 - 6t$

Set the derivative equal to zero to find the critical points:

$1 - 6t = 0$ $t = \frac{1}{6}$

Thus, $t = \frac{1}{6}$ is a critical point.

Step 2: Evaluate $f(t)$ at the critical points and endpoints

Next, we evaluate the function at the critical point $t = \frac{1}{6}$ and at the endpoints of the interval, $t = -1$ and $t = 5$.

1. $f(-1) = (-1) - 3(-1)^2 = -1 - 3 = -4$
2. $f(5) = 5 - 3(5)^2 = 5 - 75 = -70$
3. $f\left( \frac{1}{6} \right) = \frac{1}{6} - 3\left( \frac{1}{6} \right)^2 = \frac{1}{6} - 3\left( \frac{1}{36} \right) = \frac{1}{6} - \frac{1}{12} = \frac{1}{12}$

Step 3: Compare the values

Now we compare the function values:

• $f(-1) = -4$
• $f(5) = -70$
• $f\left( \frac{1}{6} \right) = \frac{1}{12}$

Step 4: Conclusion

• The absolute minimum value of $f(t)$ on the interval $[-1, 5]$ is $f(5) = -70$.
• The absolute maximum value of $f(t)$ on the interval $[-1, 5]$ is $f\left( \frac{1}{6} \right) = \frac{1}{12}$.

Final Answer:

• Absolute minimum value: $-70$
• Absolute maximum value: $\frac{1}{12}$

Would you like more details on any step? Here are a few related questions:

1. How do you compute the derivative of quadratic functions in general?
2. Why do we evaluate the function at critical points and endpoints?
3. What is the geometric meaning of the critical point in this context?
4. How do we know whether a critical point is a maximum or a minimum?
5. What would happen if the interval were unbounded?

Tip: When solving for extreme values on a closed interval, always check both the critical points and the endpoints.