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We are tasked with finding \( f(g(x)) \), where:

\[
f(x) = x^2 - 5x + 15
\]
\[
g(x) = -3x + 1
\]

To find \( f(g(x)) \), we substitute \( g(x) \) into \( f(x) \):

1. Start with \( f(x) = x^2 - 5x + 15 \), and replace every \( x \) in this expression with \( g(x) = -3x + 1 \).

So, \( f(g(x)) = ((-3x + 1)^2) - 5(-3x + 1) + 15 \).

Now, let's expand and simplify this expression:

\[
f(g(x)) = (-3x + 1)^2 - 5(-3x + 1) + 15
\]

First, expand \( (-3x + 1)^2 \):
\[
(-3x + 1)^2 = (9x^2 - 6x + 1)
\]

Next, expand \( -5(-3x + 1) \):
\[
-5(-3x + 1) = 15x - 5
\]

Now, substitute these expanded terms back into the expression:
\[
f(g(x)) = (9x^2 - 6x + 1) + (15x - 5) + 15
\]

Combine like terms:
\[
f(g(x)) = 9x^2 + (-6x + 15x) + (1 - 5 + 15)
\]
\[
f(g(x)) = 9x^2 + 9x + 11
\]

So, the final answer is:
\[
f(g(x)) = 9x^2 + 9x + 11
\]

Would you like more details or have any questions?

Here are 5 related questions to deepen understanding:
1. How would you compute \( g(f(x)) \) for the same functions?
2. What is the degree of the polynomial resulting from \( f(g(x)) \)?
3. What changes if \( f(x) \) or \( g(x) \) includes higher-degree terms?
4. How can you verify if \( f(g(x)) \) is a bijective function?
5. What is the significance of combining functions in this way in real-world applications?

**Tip**: When working with function composition, always pay close attention to substituting expressions accurately before expanding and simplifying.

Express your answer as a polynomial in standard form. Find: f(g(x)) where f(x) = x^2 - 5x + 15 and g(x) = -3x + 1.

Learn how to solve a function composition problem f(g(x)) where f(x) = x^2 - 5x + 15 and g(x) = -3x + 1. This example walks through polynomial expansion, simplifying expressions, and combining like terms, suitable for high school students.

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