Let's go through the questions step by step:

Problem 1

We are given the following functions:

• $g(x) = -2x^2 + 7x - 6$
• $f(x) = 7x^3 - 3x^2 + 4x - 6$

Part 1: Find $(f + g)(x)$

To find $(f + g)(x)$, we add the two functions: $(f + g)(x) = f(x) + g(x) = (7x^3 - 3x^2 + 4x - 6) + (-2x^2 + 7x - 6)$ Combine like terms: $(f + g)(x) = 7x^3 - 3x^2 - 2x^2 + 4x + 7x - 6 - 6$ $(f + g)(x) = 7x^3 - 5x^2 + 11x - 12$

Part 2: Find $(f - g)(1)$

Now we compute $(f - g)(x)$, which is the difference of the two functions, evaluated at $x = 1$: $(f - g)(x) = f(x) - g(x) = (7x^3 - 3x^2 + 4x - 6) - (-2x^2 + 7x - 6)$ Simplify: $(f - g)(x) = 7x^3 - 3x^2 + 4x - 6 + 2x^2 - 7x + 6$ $(f - g)(x) = 7x^3 - x^2 - 3x$ Now evaluate this at $x = 1$: $(f - g)(1) = 7(1)^3 - (1)^2 - 3(1) = 7 - 1 - 3 = 3$

Problem 2

We are given:

• $f(x) = x^3 - 216$
• $g(x) = x - 6$

Part 1: Find $(f \cdot g)(-2)$

To find $(f \cdot g)(x)$, multiply the two functions: $(f \cdot g)(x) = f(x) \cdot g(x) = (x^3 - 216)(x - 6)$ First, expand this expression: $(f \cdot g)(x) = x^3(x - 6) - 216(x - 6)$ $(f \cdot g)(x) = x^4 - 6x^3 - 216x + 1296$ Now evaluate at $x = -2$: $(f \cdot g)(-2) = (-2)^4 - 6(-2)^3 - 216(-2) + 1296$ $(f \cdot g)(-2) = 16 + 6(8) + 432 + 1296 = 16 + 48 + 432 + 1296 = 1792$

Part 2: Find $\left( \frac{f}{g} \right)(x)$

To find $\left( \frac{f}{g} \right)(x)$, divide the two functions: $\left( \frac{f}{g} \right)(x) = \frac{x^3 - 216}{x - 6}$ We can simplify this by factoring the numerator using the difference of cubes: $x^3 - 216 = (x - 6)(x^2 + 6x + 36)$ So, $\left( \frac{f}{g} \right)(x) = \frac{(x - 6)(x^2 + 6x + 36)}{x - 6}$ For $x \neq 6$, the $(x - 6)$ terms cancel, and we get: $\left( \frac{f}{g} \right)(x) = x^2 + 6x + 36$

Would you like any further clarifications or details on these steps? Here are some related questions:

1. How would you calculate $(f + g)(2)$ using the functions from Problem 1?
2. What happens if $x = 6$ in $\left( \frac{f}{g} \right)(x)$?
3. How do you factor the difference of cubes in general?
4. Can you derive the general formula for the product of two functions?
5. How would the approach change if $f(x)$ and $g(x)$ were exponential functions?

Tip: When simplifying rational expressions, always check if terms can cancel after factoring!