f(×)=2×+3,g(×)=×²-4,h(×)=×-2 tentukan (h°g°f°)(×)

Given the functions \( f(x) = 2x + 3 \), \( g(x) = x^2 - 4 \), and \( h(x) = x - 2 \), we are tasked with determining \( (h \circ g \circ f)(x) \).

This means we need to compute the composition of these functions step by step. Here's how to approach it:

### Step 1: Compute \( g(f(x)) \)
We start by finding \( g(f(x)) \), which means substituting \( f(x) \) into \( g(x) \):

\[
f(x) = 2x + 3
\]

Substitute into \( g(x) \):

\[
g(f(x)) = (2x + 3)^2 - 4
\]

Now expand \( (2x + 3)^2 \):

\[
(2x + 3)^2 = 4x^2 + 12x + 9
\]

So,

\[
g(f(x)) = 4x^2 + 12x + 9 - 4 = 4x^2 + 12x + 5
\]

### Step 2: Compute \( h(g(f(x))) \)
Now, substitute \( g(f(x)) = 4x^2 + 12x + 5 \) into \( h(x) \):

\[
h(g(f(x))) = (4x^2 + 12x + 5) - 2
\]

Simplify:

\[
h(g(f(x))) = 4x^2 + 12x + 3
\]

### Final Answer:
Thus, \( (h \circ g \circ f)(x) = 4x^2 + 12x + 3 \).

---

Do you have any questions or would you like further details on any of the steps?

Here are 5 related questions to deepen your understanding:
1. What is the general formula for composing two functions?
2. How do you expand a binomial like \( (2x + 3)^2 \)?
3. Can you apply the composition of functions to more than three functions?
4. How would the result change if \( f(x) = x - 1 \) instead of \( f(x) = 2x + 3 \)?
5. What is the inverse of \( h(x) = x - 2 \)?

**Tip**: When composing functions, always start by working from the innermost function to the outermost function.

Learn how to solve the composition of functions (h ∘ g ∘ f)(x) where f(x) = 2x + 3, g(x) = x² - 4, and h(x) = x - 2. This step-by-step solution covers key algebraic concepts such as function composition and polynomial expansion. Suitable for high school students in Grades 10-12.

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