We are given two functions:

• $f(x) = x - 8$
• $g(x) = 9x^2$

Let's go through parts (a)-(h) systematically:

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(a) $(f+g)(x)$ and its domain:

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

$(f+g)(x) = f(x) + g(x) = (x - 8) + (9x^2)$ $(f+g)(x) = 9x^2 + x - 8$

Step 2: Find the domain

Since both $f(x) = x - 8$ and $g(x) = 9x^2$ are polynomials, they are defined for all real values of $x$. Therefore, the domain is:

$\text{Domain of } (f+g)(x): (-\infty, \infty)$

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(b) $(f-g)(x)$ and its domain:

Step 1: Find $(f-g)(x)$

$(f-g)(x) = f(x) - g(x) = (x - 8) - (9x^2)$ $(f-g)(x) = -9x^2 + x - 8$

Step 2: Find the domain

Again, since $f(x)$ and $g(x)$ are polynomials, their difference is also defined for all real numbers. So, the domain is:

$\text{Domain of } (f-g)(x): (-\infty, \infty)$

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(c) $(fg)(x)$ and its domain:

Step 1: Find $(fg)(x)$

$(fg)(x) = f(x) \cdot g(x) = (x - 8) \cdot (9x^2)$ $(fg)(x) = 9x^2(x - 8) = 9x^3 - 72x^2$

Step 2: Find the domain

As both $f(x)$ and $g(x)$ are polynomials, their product is also a polynomial, which is defined for all real numbers. So, the domain is:

$\text{Domain of } (fg)(x): (-\infty, \infty)$

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(d) $\left(\frac{f}{g}\right)(x)$ and its domain:

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

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x - 8}{9x^2}$

Step 2: Find the domain

For the quotient $\frac{f(x)}{g(x)}$, we must exclude any $x$ that makes $g(x) = 9x^2 = 0$. Solving for $x$:

$9x^2 = 0 \quad \Rightarrow \quad x = 0$

Thus, the domain excludes $x = 0$. The domain is:

$\text{Domain of } \left(\frac{f}{g}\right)(x): (-\infty, 0) \cup (0, \infty)$

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(e) $f(3)$

Substitute $x = 3$ into $f(x)$: $f(3) = 3 - 8 = -5$

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(f) $g(3)$

Substitute $x = 3$ into $g(x)$: $g(3) = 9(3)^2 = 9 \times 9 = 81$

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(g) $(f+g)(3)$

Substitute $x = 3$ into $(f+g)(x) = 9x^2 + x - 8$: $(f+g)(3) = 9(3)^2 + 3 - 8 = 9 \times 9 + 3 - 8 = 81 + 3 - 8 = 76$

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(h) $\left(\frac{f}{g}\right)(3)$

Substitute $x = 3$ into $\left(\frac{f}{g}\right)(x) = \frac{x - 8}{9x^2}$: $\left(\frac{f}{g}\right)(3) = \frac{3 - 8}{9(3)^2} = \frac{-5}{9 \times 9} = \frac{-5}{81}$

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Summary:

• (a) $(f+g)(x) = 9x^2 + x - 8$, domain: $(-\infty, \infty)$
• (b) $(f-g)(x) = -9x^2 + x - 8$, domain: $(-\infty, \infty)$
• (c) $(fg)(x) = 9x^3 - 72x^2$, domain: $(-\infty, \infty)$
• (d) $\left(\frac{f}{g}\right)(x) = \frac{x - 8}{9x^2}$, domain: $(-\infty, 0) \cup (0, \infty)$
• (e) $f(3) = -5$
• (f) $g(3) = 81$
• (g) $(f+g)(3) = 76$
• (h) $\left(\frac{f}{g}\right)(3) = \frac{-5}{81}$

Would you like further explanations on any part?

Related questions:

1. What is the composition of two functions, such as $(f \circ g)(x)$?
2. How do you find the inverse of a function like $f(x) = x - 8$?
3. What are the critical points of $f(x)$ and $g(x)$?
4. How do you graph polynomial functions like $f(x)$ and $g(x)$?
5. How would you find asymptotes for rational functions like $\frac{f(x)}{g(x)}$?

Tip:

For rational functions, always check for points where the denominator equals zero to determine the domain.