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We are given the following functions:

- \( f(x) = \sqrt{x^2 - 2x + 6} \)
- \( g(x) = 7x + 3 \)

### 1. Finding \( (f \circ g)(x) \):
The composition \( (f \circ g)(x) \) means we need to substitute \( g(x) \) into \( f(x) \), so:

\[
(f \circ g)(x) = f(g(x)) = f(7x + 3)
\]

Substitute \( 7x + 3 \) into the expression for \( f(x) \):

\[
f(7x + 3) = \sqrt{(7x + 3)^2 - 2(7x + 3) + 6}
\]

Now, simplify step by step:

- First, expand \( (7x + 3)^2 \):
  \[
  (7x + 3)^2 = 49x^2 + 42x + 9
  \]
- Next, expand \( -2(7x + 3) \):
  \[
  -2(7x + 3) = -14x - 6
  \]
- Now, substitute these back into the square root:
  \[
  f(7x + 3) = \sqrt{49x^2 + 42x + 9 - 14x - 6 + 6}
  \]
  Simplifying the terms:
  \[
  f(7x + 3) = \sqrt{49x^2 + 28x + 9}
  \]

### 2. Finding \( (f \circ g)(-3) \):
Now, substitute \( x = -3 \) into \( (f \circ g)(x) \):

\[
(f \circ g)(-3) = \sqrt{49(-3)^2 + 28(-3) + 9}
\]
First, calculate each term:

\[
49(-3)^2 = 49(9) = 441
\]
\[
28(-3) = -84
\]
\[
441 - 84 + 9 = 366
\]

So:

\[
(f \circ g)(-3) = \sqrt{366}
\]

Thus, \( (f \circ g)(-3) \approx 19.13 \) (since \( \sqrt{366} \approx 19.13 \)).

---

Would you like any additional details or clarification?

Here are 5 related questions for further exploration:

1. How do we verify the domain of the composite function \( f \circ g \)?
2. What are the steps for calculating the derivative of \( f(x) = \sqrt{x^2 - 2x + 6} \)?
3. How would you evaluate \( (f \circ g)(x) \) for a different value, say \( x = 2 \)?
4. How do the graphs of \( f(x) \), \( g(x) \), and \( f \circ g(x) \) compare?
5. What if \( g(x) \) was a quadratic function instead of a linear one?

**Tip:** When performing function composition, ensure the input into the outer function is valid for the function's domain!

Suppose f(x) = √(x² - 2x + 6) and g(x) = 7x + 3. Find (f ∘ g)(x) and (f ∘ g)(-3).

Learn how to solve a function composition problem involving f(x) = √(x² - 2x + 6) and g(x) = 7x + 3. Step-by-step explanation provided for finding (f ∘ g)(x) and evaluating (f ∘ g)(-3). Suitable for high school students.

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