Let's solve the given problems step-by-step. The functions provided are:

$f(x) = x - 9$ $g(x) = 6x^2$

(e) Find $(f + g)(3)$:

$(f + g)(x) = f(x) + g(x) = (x - 9) + 6x^2$

Plugging in $x = 3$: $(f + g)(3) = (3 - 9) + 6 \times 3^2$ $(f + g)(3) = -6 + 6 \times 9$ $(f + g)(3) = -6 + 54 = 48$

Answer for (e): $(f + g)(3) = 48$

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

$(f - g)(x) = f(x) - g(x) = (x - 9) - 6x^2$

Plugging in $x = 2$: $(f - g)(2) = (2 - 9) - 6 \times 2^2$ $(f - g)(2) = -7 - 6 \times 4$ $(f - g)(2) = -7 - 24 = -31$

Answer for (f): $(f - g)(2) = -31$

(g) Find $(f \cdot g)(4)$:

$(f \cdot g)(x) = f(x) \cdot g(x) = (x - 9) \cdot 6x^2$

Plugging in $x = 4$: $(f \cdot g)(4) = (4 - 9) \cdot 6 \times 4^2$ $(f \cdot g)(4) = -5 \cdot 6 \times 16$ $(f \cdot g)(4) = -5 \cdot 96 = -480$

Answer for (g): $(f \cdot g)(4) = -480$

(h) Find $\left(\frac{f}{g}\right)(1)$:

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x - 9}{6x^2}$

Plugging in $x = 1$: $\left(\frac{f}{g}\right)(1) = \frac{1 - 9}{6 \times 1^2}$ $\left(\frac{f}{g}\right)(1) = \frac{-8}{6}$ $\left(\frac{f}{g}\right)(1) = -\frac{4}{3}$

Answer for (h): $\left(\frac{f}{g}\right)(1) = -\frac{4}{3}$

Do you need any more details or have any questions? Here are some related questions:

1. What is the domain of the function $\left(\frac{f}{g}\right)(x)$?
2. How would you find $(f + g)(x)$ in its simplified form?
3. What is the value of $(f - g)(x)$ when $x = 0$?
4. Can you find the zeros of the function $g(x)$?
5. What would be the derivative of $f(x)$ and $g(x)$?

Tip: Always verify the domain of your function, especially when dealing with divisions to avoid undefined values.