Let's go through each part of the question based on the given functions:

• $f(x) = x^2 - 3$
• $g(x) = 4x + 1$

Now, let's solve each part.

Part A: $(f + g)(x)$

$(f + g)(x) = f(x) + g(x)$ $= (x^2 - 3) + (4x + 1)$ $= x^2 + 4x - 2$

Part B: $(f - g)(x)$

$(f - g)(x) = f(x) - g(x)$ $= (x^2 - 3) - (4x + 1)$ $= x^2 - 4x - 4$

Part C: $(f \cdot g)(x)$

$(f \cdot g)(x) = f(x) \cdot g(x)$ $= (x^2 - 3)(4x + 1)$ Expanding this: $= x^2 \cdot 4x + x^2 \cdot 1 - 3 \cdot 4x - 3 \cdot 1$ $= 4x^3 + x^2 - 12x - 3$

Part D: $\left(\frac{f}{g}\right)(x)$

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$ $= \frac{x^2 - 3}{4x + 1}$

Part E: $(f \circ g)(x)$ — Composition of $f$ and $g$

$(f \circ g)(x) = f(g(x))$ $= f(4x + 1)$ Since $f(x) = x^2 - 3$, substitute $4x + 1$ for $x$ in $f(x)$: $= (4x + 1)^2 - 3$ Expanding $(4x + 1)^2$: $= 16x^2 + 8x + 1 - 3$ $= 16x^2 + 8x - 2$

Part G: $(f \circ g)(2)$

1. First, find $g(2)$: $g(2) = 4(2) + 1 = 8 + 1 = 9$
2. Now, find $f(g(2)) = f(9)$: $f(9) = 9^2 - 3 = 81 - 3 = 78$

Summary of Answers:

• A) $(f + g)(x) = x^2 + 4x - 2$
• B) $(f - g)(x) = x^2 - 4x - 4$
• C) $(f \cdot g)(x) = 4x^3 + x^2 - 12x - 3$
• D) $\left(\frac{f}{g}\right)(x) = \frac{x^2 - 3}{4x + 1}$
• E) $(f \circ g)(x) = 16x^2 + 8x - 2$
• G) $(f \circ g)(2) = 78$

Would you like further details or explanations for any of these solutions?

1. How would the results change if $f(x)$ and $g(x)$ were different?
2. How do you find the composition of two functions in general?
3. What is the difference between function addition and composition?
4. Why do we use parentheses in notation like $(f + g)(x)$ and $(f \circ g)(x)$?
5. What are some applications of function operations in real-world problems?

Tip: When finding compositions, always substitute carefully to avoid common mistakes in simplification.