To find the instantaneous rate of change of the function \( f(t) = \frac{5}{t} - 1 \) at \( t = -1 \), we need to compute the derivative \( f'(t) \), then evaluate it at \( t = -1 \).

### Step 1: Differentiate \( f(t) \)
We use the power rule and the derivative of a constant to differentiate the function.

\[
f(t) = \frac{5}{t} - 1 = 5t^{-1} - 1
\]

The derivative of \( 5t^{-1} \) is:

\[
f'(t) = -5t^{-2}
\]

The derivative of a constant \( -1 \) is zero, so we have:

\[
f'(t) = -\frac{5}{t^2}
\]

### Step 2: Evaluate the derivative at \( t = -1 \)

Now, substitute \( t = -1 \) into the derivative:

\[
f'(-1) = -\frac{5}{(-1)^2} = -\frac{5}{1} = -5
\]

Thus, the instantaneous rate of change of the function at \( t = -1 \) is \( -5 \).

### Summary:
The instantaneous rate of change of \( f(t) = \frac{5}{t} - 1 \) at \( t = -1 \) is \( -5 \).

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Would you like more details on the differentiation process, or do you have any questions?

Here are 5 related questions you may find helpful:
1. How do you compute the average rate of change of a function between two points?
2. What is the physical meaning of an instantaneous rate of change?
3. How do you interpret the derivative in real-life contexts?
4. Can you find the instantaneous rate of change for more complex functions?
5. How does the power rule work in derivative calculations?

**Tip:** When taking the derivative of a function involving a fraction, rewriting it with a negative exponent often makes differentiation easier!

Learn how to compute the instantaneous rate of change of the function f(t) = 5/t - 1 at t = -1 using differentiation. This problem covers key calculus concepts like the derivative and power rule, suitable for students in Grades 10-12.

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