Let's break down each part of your question step by step.

Given Functions:

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

B. Find $(g \circ f)(x)$:

The composition $(g \circ f)(x)$ means we are plugging $f(x)$ into $g(x)$. This is the same as:

$(g \circ f)(x) = g(f(x))$

Substitute $f(x) = 3 - x$ into $g(x)$:

$g(f(x)) = g(3 - x)$

Now, substitute $3 - x$ into $g(x)$:

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

Let's expand this expression:

$(3 - x)^2 = 9 - 6x + x^2$ $g(3 - x) = 2(9 - 6x + x^2) + (3 - x) + 4$ $g(3 - x) = 2(9) - 2(6x) + 2(x^2) + 3 - x + 4$ $g(3 - x) = 18 - 12x + 2x^2 + 3 - x + 4$ $g(3 - x) = 2x^2 - 13x + 25$

So, $(g \circ f)(x) = 2x^2 - 13x + 25$.

C. Find $(f \circ g)(2)$:

The composition $(f \circ g)(2)$ means we need to first evaluate $g(2)$, and then apply $f$ to that result. So we do the following:

1. Evaluate $g(2)$: $g(2) = 2(2)^2 + 2 + 4$ $g(2) = 2(4) + 2 + 4 = 8 + 2 + 4 = 14$

2. Now, apply $f$ to $g(2) = 14$: $f(14) = 3 - 14 = -11$

So, $(f \circ g)(2) = -11$.

D. Find $(g \circ f)(2)$:

To find $(g \circ f)(2)$, we first need to evaluate $f(2)$, and then apply $g$ to that result.

1. Evaluate $f(2)$: $f(2) = 3 - 2 = 1$

2. Now, apply $g$ to $f(2) = 1$: $g(1) = 2(1)^2 + 1 + 4 = 2 + 1 + 4 = 7$

So, $(g \circ f)(2) = 7$.

Summary of Results:

• $(g \circ f)(x) = 2x^2 - 13x + 25$
• $(f \circ g)(2) = -11$
• $(g \circ f)(2) = 7$

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

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Here are 5 related questions to explore further:

1. How do you find the composition of functions in general?
2. What would happen if we had more complex functions for $f(x)$ and $g(x)$?
3. How can the composition of functions be used in real-world scenarios?
4. What is the difference between $(g \circ f)(x)$ and $(f \circ g)(x)$?
5. How do you handle compositions involving higher-degree polynomials?

Tip: When composing functions, always be careful to substitute the entire expression of one function into the other and simplify step by step.