To find the absolute maximum and minimum values of the function $f(t) = t - \frac{3}{t}$ on the interval $[-1, 5]$, we’ll follow these steps:

Step 1: Find $f'(t)$

First, we find the derivative $f'(t)$ to identify any critical points in the interval.

$f(t) = t - \frac{3}{t}$

Using the power rule and the derivative of $t^{-1}$, we get:

$f'(t) = 1 + \frac{3}{t^2}$

Step 2: Set $f'(t) = 0$ to Find Critical Points

Now we set the derivative equal to zero and solve for $t$:

$1 + \frac{3}{t^2} = 0$

Subtract 1 from both sides:

$\frac{3}{t^2} = -1$

This equation has no real solutions because $\frac{3}{t^2}$ is always positive for real $t$. Therefore, there are no critical points in the interval that we obtain by setting $f'(t) = 0$.

Step 3: Evaluate $f(t)$ at the Endpoints of the Interval

Since there are no critical points in $[-1, 5]$, we only need to evaluate $f(t)$ at the endpoints of the interval.

1. At $t = -1$:

   $f(-1) = -1 - \frac{3}{-1} = -1 + 3 = 2$

2. At $t = 5$:

   $f(5) = 5 - \frac{3}{5} = 5 - 0.6 = 4.4$

Step 4: Determine the Absolute Maximum and Minimum

From the evaluations above:

• $f(-1) = 2$
• $f(5) = 4.4$

Thus:

• The absolute minimum value of $f$ on $[-1, 5]$ is $f(-1) = 2$.
• The absolute maximum value of $f$ on $[-1, 5]$ is $f(5) = 4.4$.

Final Answer

• Absolute minimum: $f(-1) = 2$
• Absolute maximum: $f(5) = 4.4$

Would you like additional details, or do you have questions about any part of this process?

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1. How do we interpret critical points when finding extrema on closed intervals?
2. What is the significance of evaluating endpoints in finding absolute extrema?
3. How does the function behavior influence the existence of critical points?
4. What techniques can be used when no critical points satisfy the derivative equation?
5. How does analyzing the derivative help in determining concavity and monotonicity?

Tip: Always check the endpoints of a closed interval when finding absolute extrema, as they can often provide the maximum or minimum values.