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We are given two functions:
- \( f(x) = -3x - 1 \)
- \( h(x) = -\frac{6}{x+1} \)

We are tasked with finding \( (f \circ h)(-3) \), which means we first need to evaluate \( h(-3) \) and then substitute that result into \( f(x) \).

### Step 1: Evaluate \( h(-3) \)

\[
h(-3) = -\frac{6}{-3+1} = -\frac{6}{-2} = 3
\]

### Step 2: Evaluate \( f(h(-3)) = f(3) \)

\[
f(3) = -3(3) - 1 = -9 - 1 = -10
\]

Thus, \( (f \circ h)(-3) = -10 \).

Would you like any further explanation?

Here are 5 related questions to expand on this concept:
1. What is the value of \( (h \circ f)(-3) \)?
2. How do you determine the domain of \( f(x) \) and \( h(x) \)?
3. Can you find \( (f \circ h)(0) \)?
4. What happens when you compose \( h \) with itself, i.e., \( (h \circ h)(x) \)?
5. How would you graph the composition \( (f \circ h)(x) \)?

**Tip**: When working with function compositions, always evaluate the inner function first before substituting its result into the outer function.

Let f(x) = -3x - 1, h(x) = -6/(x + 1). Find (f ∘ h)(-3).

Learn how to solve the function composition problem with f(x) = -3x - 1 and h(x) = -6/(x + 1), focusing on algebraic manipulations and evaluating at specific points like x = -3. Suitable for students in Grades 9-11.

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