Let's solve the problem step by step.

We are given the functions:

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

We need to solve for the following:

a. $(f \circ g)(x)$

This is the composition $f(g(x))$, which means we substitute $g(x)$ into $f(x)$.

$f(g(x)) = 2 - g(x) = 2 - (3x^2 + x + 3)$ Simplifying: $f(g(x)) = 2 - 3x^2 - x - 3 = -3x^2 - x - 1$

Thus, $(f \circ g)(x) = -3x^2 - x - 1$.

b. $(g \circ f)(x)$

This is the composition $g(f(x))$, which means we substitute $f(x)$ into $g(x)$.

$g(f(x)) = 3(f(x))^2 + f(x) + 3$ Substituting $f(x) = 2 - x$: $g(f(x)) = 3(2 - x)^2 + (2 - x) + 3$ First, expand $(2 - x)^2$: $(2 - x)^2 = 4 - 4x + x^2$ Now substitute into the equation: $g(f(x)) = 3(4 - 4x + x^2) + (2 - x) + 3$ Distribute the 3: $g(f(x)) = 12 - 12x + 3x^2 + 2 - x + 3$ Simplify: $g(f(x)) = 3x^2 - 13x + 17$

Thus, $(g \circ f)(x) = 3x^2 - 13x + 17$.

c. $(f \circ g)(2)$

We already found $(f \circ g)(x)$ to be $-3x^2 - x - 1$, so now we substitute $x = 2$:

$f(g(2)) = -3(2)^2 - (2) - 1 = -3(4) - 2 - 1 = -12 - 2 - 1 = -15$

Thus, $(f \circ g)(2) = -15$.

d. $(g \circ f)(2)$

We already found $(g \circ f)(x)$ to be $3x^2 - 13x + 17$, so now we substitute $x = 2$:

$g(f(2)) = 3(2)^2 - 13(2) + 17 = 3(4) - 26 + 17 = 12 - 26 + 17 = 3$

Thus, $(g \circ f)(2) = 3$.

Final Answers:

• a. $(f \circ g)(x) = -3x^2 - x - 1$
• b. $(g \circ f)(x) = 3x^2 - 13x + 17$
• c. $(f \circ g)(2) = -15$
• d. $(g \circ f)(2) = 3$

Would you like further explanations or have any questions?

Here are 5 related questions to expand your knowledge:

1. How do you interpret the domain of a composed function?
2. Can you find $(f \circ f)(x)$ and $(g \circ g)(x)$?
3. What is the difference between function composition and regular multiplication of functions?
4. How does the order of composition affect the result of two functions?
5. How do you differentiate a composed function using the chain rule?

Tip: When composing functions, make sure to carefully substitute one function into the other, and always simplify the expression step by step to avoid errors.