For ​f(x)equalsnothingxminus3 and ​g(x)equals5xsquaredminus2​, find the following functions.
a. left parenthesis f circle g right parenthesis left parenthesis x right parenthesis​; b. left parenthesis g circle f right parenthesis left parenthesis x right parenthesis​; c. left parenthesis f circle g right parenthesis left parenthesis 1 right parenthesis​; d. left parenthesis g circle f right parenthesis left parenthesis 1 right parenthesis


Given functions:
- \( f(x) = x - 3 \)
- \( g(x) = 5x^2 - 2 \)

We want to find:
1. \( (f \circ g)(x) \)
2. \( (g \circ f)(x) \)
3. \( (f \circ g)(1) \)
4. \( (g \circ f)(1) \)

Let's solve each part step-by-step.

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### Part (a): \( (f \circ g)(x) \)
The notation \( (f \circ g)(x) \) represents the composite function \( f(g(x)) \), where we substitute \( g(x) \) into \( f(x) \).

1. Substitute \( g(x) = 5x^2 - 2 \) into \( f(x) \):
   \[
   f(g(x)) = f(5x^2 - 2)
   \]
2. Since \( f(x) = x - 3 \), we replace \( x \) in \( f(x) \) with \( 5x^2 - 2 \):
   \[
   f(5x^2 - 2) = (5x^2 - 2) - 3
   \]
3. Simplify:
   \[
   f(g(x)) = 5x^2 - 5
   \]

Thus, \( (f \circ g)(x) = 5x^2 - 5 \).

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### Part (b): \( (g \circ f)(x) \)
The notation \( (g \circ f)(x) \) represents the composite function \( g(f(x)) \), where we substitute \( f(x) \) into \( g(x) \).

1. Substitute \( f(x) = x - 3 \) into \( g(x) \):
   \[
   g(f(x)) = g(x - 3)
   \]
2. Since \( g(x) = 5x^2 - 2 \), we replace \( x \) in \( g(x) \) with \( x - 3 \):
   \[
   g(x - 3) = 5(x - 3)^2 - 2
   \]
3. Expand \( (x - 3)^2 \):
   \[
   (x - 3)^2 = x^2 - 6x + 9
   \]
4. Substitute back:
   \[
   g(x - 3) = 5(x^2 - 6x + 9) - 2
   \]
5. Distribute and simplify:
   \[
   g(x - 3) = 5x^2 - 30x + 45 - 2
   \]
   \[
   g(x - 3) = 5x^2 - 30x + 43
   \]

Thus, \( (g \circ f)(x) = 5x^2 - 30x + 43 \).

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### Part (c): \( (f \circ g)(1) \)
Now, we substitute \( x = 1 \) into \( (f \circ g)(x) \).

1. First, find \( g(1) \):
   \[
   g(1) = 5(1)^2 - 2 = 5 - 2 = 3
   \]
2. Then, substitute \( g(1) = 3 \) into \( f(x) \):
   \[
   f(g(1)) = f(3)
   \]
3. Calculate \( f(3) \):
   \[
   f(3) = 3 - 3 = 0
   \]

Thus, \( (f \circ g)(1) = 0 \).

---

### Part (d): \( (g \circ f)(1) \)
Finally, substitute \( x = 1 \) into \( (g \circ f)(x) \).

1. First, find \( f(1) \):
   \[
   f(1) = 1 - 3 = -2
   \]
2. Then, substitute \( f(1) = -2 \) into \( g(x) \):
   \[
   g(f(1)) = g(-2)
   \]
3. Calculate \( g(-2) \):
   \[
   g(-2) = 5(-2)^2 - 2 = 5 \cdot 4 - 2 = 20 - 2 = 18
   \]

Thus, \( (g \circ f)(1) = 18 \).

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### Summary of Results
- **(a)** \( (f \circ g)(x) = 5x^2 - 5 \)
- **(b)** \( (g \circ f)(x) = 5x^2 - 30x + 43 \)
- **(c)** \( (f \circ g)(1) = 0 \)
- **(d)** \( (g \circ f)(1) = 18 \)

Would you like any details on these steps or have questions?

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### Further Related Questions
1. What is the domain of \( (f \circ g)(x) \)?
2. What is the range of \( (f \circ g)(x) \)?
3. How would the composition change if \( f(x) \) were a different linear function, like \( f(x) = 2x + 1 \)?
4. Can composite functions be commutative for all functions?
5. How do you find the inverse of \( f(g(x)) \)?

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### Tip:
When forming composite functions, always work step-by-step to substitute and simplify, ensuring each inner function is fully evaluated before moving to the next.

For f(x) = x - 3 and g(x) = 5x^2 - 2, find (f ∘ g)(x), (g ∘ f)(x), (f ∘ g)(1), and (g ∘ f)(1).

Explore the detailed solutions for the composition of functions f(x) = x - 3 and g(x) = 5x^2 - 2, including (f ∘ g)(x), (g ∘ f)(x), (f ∘ g)(1), and (g ∘ f)(1). This page covers important concepts in algebra and function composition, suitable for Grades 9-10.

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