Let's solve the given problem step by step.

Given Functions:

• $f(x) = 10x - 9$
• $g(x) = \sqrt{x^2 - 6x + 10}$

We need to find:

1. $(f \circ g)(x)$, which means $f(g(x))$
2. $(f \circ g)(-4)$

1. Finding $(f \circ g)(x)$:

$f(g(x))$ means we substitute $g(x)$ into $f(x)$.

We know:

• $f(x) = 10x - 9$
• $g(x) = \sqrt{x^2 - 6x + 10}$

So, $f(g(x)) = 10 \cdot g(x) - 9$.

Substitute $g(x)$: $f(g(x)) = 10 \cdot \sqrt{x^2 - 6x + 10} - 9$

Thus, $(f \circ g)(x) = 10 \cdot \sqrt{x^2 - 6x + 10} - 9$

2. Finding $(f \circ g)(-4)$:

We substitute $x = -4$ into $g(x)$ first, and then find $f(g(-4))$.

First, calculate $g(-4)$: $g(-4) = \sqrt{(-4)^2 - 6(-4) + 10} = \sqrt{16 + 24 + 10} = \sqrt{50}$ So, $g(-4) = \sqrt{50}$, which simplifies to $g(-4) = 5\sqrt{2}$.

Now, substitute $g(-4) = 5\sqrt{2}$ into $f(x)$: $f(g(-4)) = 10 \cdot 5\sqrt{2} - 9 = 50\sqrt{2} - 9$

Thus, $(f \circ g)(-4) = 50\sqrt{2} - 9$

Final Results:

1. $(f \circ g)(x) = 10 \cdot \sqrt{x^2 - 6x + 10} - 9$
2. $(f \circ g)(-4) = 50\sqrt{2} - 9$

Would you like more details on any part of this, or do you have further questions?

Here are some related questions to expand on the topic:

1. How do you find the domain of a composite function?
2. What is the difference between $f(g(x))$ and $g(f(x))$?
3. How do you handle the square root function when working with domains?
4. Can you calculate the range of $(f \circ g)(x)$?
5. How does function composition affect the graph of the resulting function?

Tip: Always check the domain of the inner function before composing two functions, especially when dealing with square roots or fractions!