To solve the given problem, we need to find $(g \circ f)(-2 + x)$, which means we first need to compute $f(-2 + x)$ and then use that result as the input for $g(x)$.

Given functions:

• $g(x) = 2x - 2$
• $f(x) = x^2 + 3x$

Step 1: Find $f(-2 + x)$

We substitute $(-2 + x)$ into the function $f(x)$:

$f(-2 + x) = (-2 + x)^2 + 3(-2 + x)$

Expanding this expression:

$(-2 + x)^2 = 4 - 4x + x^2$ $3(-2 + x) = -6 + 3x$

Now, combine both parts:

$f(-2 + x) = (4 - 4x + x^2) + (-6 + 3x)$ $f(-2 + x) = x^2 - x - 2$

Step 2: Use this result in $g(x)$

Now that we have $f(-2 + x) = x^2 - x - 2$, we substitute this into $g(x)$:

$g(f(-2 + x)) = g(x^2 - x - 2)$ Substitute into $g(x)$:

$g(x^2 - x - 2) = 2(x^2 - x - 2) - 2$ Expanding this:

$g(x^2 - x - 2) = 2x^2 - 2x - 4 - 2$ $g(x^2 - x - 2) = 2x^2 - 2x - 6$

Thus, the final answer is:

$(g \circ f)(-2 + x) = 2x^2 - 2x - 6$

Would you like any more details or explanations? Here are five related questions to deepen your understanding:

1. What is the composition of two functions and how does it work in general?
2. How do you expand a binomial like $(-2 + x)^2$?
3. What happens when you compose two functions in the reverse order, such as $(f \circ g)(x)$?
4. Can the process of function composition change the domain of the resulting function?
5. What are the potential real-life applications of function composition?

Tip: Always be mindful of the order when composing functions; mixing up the order can change the result!