Let's go through each of the sub-problems step by step.

We are given two functions: $f(x) = x^2 - 2 \quad \text{and} \quad g(x) = 3x - 2$

Part a: $f(-2)$

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

$f(-2) = (-2)^2 - 2 = 4 - 2 = 2$

Thus, $f(-2) = 2$.

Part b: $f(-1) + g(-2)$

1. First, calculate $f(-1)$: $f(-1) = (-1)^2 - 2 = 1 - 2 = -1$

2. Now, calculate $g(-2)$: $g(-2) = 3(-2) - 2 = -6 - 2 = -8$

Now, add the two results: $f(-1) + g(-2) = -1 + (-8) = -9$

Thus, $f(-1) + g(-2) = -9$.

Part c: $3f(x) + g(x)$

We substitute the expressions for $f(x)$ and $g(x)$:

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

Now, expand and simplify:

$3f(x) + g(x) = 3x^2 - 6 + 3x - 2 = 3x^2 + 3x - 8$

Thus, $3f(x) + g(x) = 3x^2 + 3x - 8$.

Part d: $f(g(4))$

1. First, calculate $g(4)$: $g(4) = 3(4) - 2 = 12 - 2 = 10$

2. Now, substitute this result into $f(x)$: $f(10) = (10)^2 - 2 = 100 - 2 = 98$

Thus, $f(g(4)) = 98$.

Part e: $g(f(4))$

1. First, calculate $f(4)$: $f(4) = (4)^2 - 2 = 16 - 2 = 14$

2. Now, substitute this result into $g(x)$: $g(14) = 3(14) - 2 = 42 - 2 = 40$

Thus, $g(f(4)) = 40$.

Final Answers:

• a. $f(-2) = 2$
• b. $f(-1) + g(-2) = -9$
• c. $3f(x) + g(x) = 3x^2 + 3x - 8$
• d. $f(g(4)) = 98$
• e. $g(f(4)) = 40$

Would you like any further details or explanations?

Here are five related questions for further practice:

1. What is $f(g(-1))$?
2. Find the value of $g(f(3))$.
3. Simplify $2f(x) + g(x)$.
4. What is $f(x) + g(x)$ in general form?
5. Calculate $f(g(0))$.

Tip: When evaluating composite functions, always start from the inside and work outward!