Evaluate 
h(3),
 where 
h = g ∘ f.
f(x) = 
3

x2 − 9
,    g(x) = 5x3 + 9
h(3) =

We are given two functions \( f(x) = 3x^2 - 9 \) and \( g(x) = 5x^3 + 9 \), and we are asked to evaluate \( h(3) \), where \( h = g \circ f \). This means that \( h(x) = g(f(x)) \), i.e., we first apply \( f(x) \), and then apply \( g(x) \) to the result of \( f(x) \).

Let's calculate \( h(3) \) step by step:

### Step 1: Compute \( f(3) \)
Using the function \( f(x) = 3x^2 - 9 \), substitute \( x = 3 \):

\[
f(3) = 3(3)^2 - 9 = 3(9) - 9 = 27 - 9 = 18
\]

### Step 2: Compute \( g(f(3)) = g(18) \)
Now, using the function \( g(x) = 5x^3 + 9 \), substitute \( x = 18 \):

\[
g(18) = 5(18)^3 + 9
\]

First, calculate \( 18^3 \):

\[
18^3 = 18 \times 18 \times 18 = 5832
\]

Now compute \( g(18) \):

\[
g(18) = 5(5832) + 9 = 29160 + 9 = 29169
\]

### Final Answer:
Therefore, \( h(3) = 29169 \).

Would you like more details or have any questions? Here are some related questions you might explore:

1. What is \( h(2) \) if \( h = g \circ f \)?
2. How does the composition of two functions affect their domain?
3. Can you simplify \( h(x) = g(f(x)) \) in general form?
4. What happens if you reverse the composition, \( h = f \circ g \)?
5. How do transformations of functions affect their graphs?

**Tip**: When composing functions, always remember to evaluate the inner function first, then apply the outer function.

Evaluate h(3), where h = g ∘ f. f(x) = 3x^2 − 9, g(x) = 5x^3 + 9

Evaluate h(3) using function composition where h = g ∘ f, f(x) = 3x^2 − 9, and g(x) = 5x^3 + 9. Follow this step-by-step solution to understand how polynomials and function composition work together. Suitable for students in Grades 10-12.

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