To find the instantaneous rate of change of $h(t) = \frac{3}{t+3}$ at the point $t = -2$, we need to calculate the derivative of $h(t)$ with respect to $t$ and evaluate it at $t = -2$.

1. Define the function: $h(t) = \frac{3}{t + 3}$

2. Differentiate $h(t)$ with respect to $t$: Using the quotient rule, where $h(t) = \frac{f(t)}{g(t)}$ with $f(t) = 3$ and $g(t) = t + 3$, the derivative $h'(t)$ is: $h'(t) = \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2}$ Since $f(t) = 3$ (a constant), $f'(t) = 0$, and $g(t) = t + 3$ so $g'(t) = 1$, we have: $h'(t) = \frac{0 \cdot (t + 3) - 3 \cdot 1}{(t + 3)^2} = \frac{-3}{(t + 3)^2}$

3. Evaluate $h'(t)$ at $t = -2$: Substitute $t = -2$ into the derivative: $h'(-2) = \frac{-3}{(-2 + 3)^2} = \frac{-3}{1} = -3$

Thus, the instantaneous rate of change of $h(t)$ at $t = -2$ is $\mathbf{-3}$.

Would you like further details, or do you have any questions?

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Here are some related questions you might explore:

1. How would you interpret the meaning of this rate of change?
2. What is the instantaneous rate of change at other points for $h(t)$?
3. How does the rate of change of $h(t)$ compare to a linear function?
4. Could you use limits to verify the instantaneous rate of change?
5. How would you find the instantaneous rate of change at any arbitrary point $t = a$?

Tip: For rational functions like $\frac{3}{t+3}$, the quotient rule can streamline finding the derivative efficiently!