Math worksheet for practicing function operations and domain analysis.

Problem: Perform the indicated operation and state the domain for each problem.

#### Problem 1:
Given:
\[
g(c) = 3c + 5, \quad h(c) = c^2 + 2
\]
Find \((g \cdot h)(c)\).

Solution:
The operation \( (g \cdot h)(c) \) means we need to multiply the functions \( g(c) \) and \( h(c) \):
\[
(g \cdot h)(c) = g(c) \cdot h(c)
\]
Substitute the given functions:
\[
g(c) = 3c + 5, \quad h(c) = c^2 + 2
\]
\[
(g \cdot h)(c) = (3c + 5)(c^2 + 2)
\]
Now, expand the product using the distributive property:
\[
(g \cdot h)(c) = 3c \cdot c^2 + 3c \cdot 2 + 5 \cdot c^2 + 5 \cdot 2
\]
\[
(g \cdot h)(c) = 3c^3 + 6c + 5c^2 + 10
\]
Combine like terms (though there are no like terms here):
\[
(g \cdot h)(c) = 3c^3 + 5c^2 + 6c + 10
\]

Domain:
Both \( g(c) \) and \( h(c) \) are polynomial functions, and polynomial functions are defined for all real numbers. Therefore, the domain of \( (g \cdot h)(c) \) is:
\[
\boxed{(-\infty, \infty)}
\]

---

#### Problem 2:
Given:
\[
p(x) = \sqrt{x - 3}, \quad b(x) = \frac{x^2 + x - 2}{x - 1}
\]
Find \(\frac{p(x)}{b(x)}\).

Solution:
The operation \(\frac{p(x)}{b(x)}\) means we need to divide the function \( p(x) \) by the function \( b(x) \):
\[
\frac{p(x)}{b(x)} = \frac{\sqrt{x - 3}}{\frac{x^2 + x - 2}{x - 1}}
\]
Simplify the division by multiplying by the reciprocal of \( b(x) \):
\[
\frac{p(x)}{b(x)} = \sqrt{x - 3} \cdot \frac{x - 1}{x^2 + x - 2}
\]
Next, factor the quadratic in the denominator of \( b(x) \):
\[
x^2 + x - 2 = (x + 2)(x - 1)
\]
So,
\[
b(x) = \frac{(x + 2)(x - 1)}{x - 1}
\]
For \( x \neq 1 \), the \( x - 1 \) terms cancel:
\[
b(x) = x + 2 \quad \text{(for } x \neq 1\text{)}
\]
Thus,
\[
\frac{p(x)}{b(x)} = \frac{\sqrt{x - 3}}{x + 2}
\]

Domain:
1. For \( p(x) = \sqrt{x - 3} \), the expression inside the square root must be non-negative:
\[
x - 3 \geq 0 \implies x \geq 3
\]
2. For \( b(x) = x + 2 \), the denominator cannot be zero:
\[
x + 2 \neq 0 \implies x \neq -2
\]
3. Additionally, since \( b(x) \) was originally defined as a rational function with \( x - 1 \) in the denominator, we must exclude \( x = 1 \). However, this is already covered by the condition \( x \geq 3 \).

Combining these conditions, the domain is:
\[
x \geq 3
\]
In interval notation:
\[
\boxed{[3, \infty)}
\]

---

#### Problem 3:
Given:
\[
f(b) = b - 5, \quad g(b) = b^2 + b
\]
Find \( g(b) - f(b) \).

Solution:
The operation \( g(b) - f(b) \) means we need to subtract the function \( f(b) \) from \( g(b) \):
\[
g(b) - f(b) = (b^2 + b) - (b - 5)
\]
Distribute the negative sign:
\[
g(b) - f(b) = b^2 + b - b + 5
\]
Combine like terms:
\[
g(b) - f(b) = b^2 + 5
\]

Domain:
Both \( f(b) \) and \( g(b) \) are polynomial functions, and polynomial functions are defined for all real numbers. Therefore, the domain of \( g(b) - f(b) \) is:
\[
\boxed{(-\infty, \infty)}
\]

---

#### Problem 4:
Given:
\[
h(k) = 4k - 3, \quad m(k) = 4k - 5
\]
Find \( (3h - 5m)(k) \).

Solution:
The operation \( (3h - 5m)(k) \) means we need to compute \( 3h(k) - 5m(k) \):
\[
(3h - 5m)(k) = 3h(k) - 5m(k)
\]
Substitute the given functions:
\[
h(k) = 4k - 3, \quad m(k) = 4k - 5
\]
\[
3h(k) = 3(4k - 3) = 12k - 9
\]
\[
5m(k) = 5(4k - 5) = 20k - 25
\]
Now, compute \( 3h(k) - 5m(k) \):
\[
(3h - 5m)(k) = (12k - 9) - (20k - 25)
\]
Distribute the negative sign:
\[
(3h - 5m)(k) = 12k - 9 - 20k + 25
\]
Combine like terms:
\[
(3h - 5m)(k) = (12k - 20k) + (-9 + 25)
\]
\[
(3h - 5m)(k) = -8k + 16
\]

Domain:
Both \( h(k) \) and \( m(k) \) are linear functions, and linear functions are defined for all real numbers. Therefore, the domain of \( (3h - 5m)(k) \) is:
\[
\boxed{(-\infty, \infty)}
\]

---

Final Answers:

1. \((g \cdot h)(c) = 3c^3 + 5c^2 + 6c + 10\), Domain: \(\boxed{(-\infty, \infty)}\)
2. \(\frac{p(x)}{b(x)} = \frac{\sqrt{x - 3}}{x + 2}\), Domain: \(\boxed{[3, \infty)}\)
3. \(g(b) - f(b) = b^2 + 5\), Domain: \(\boxed{(-\infty, \infty)}\)
4. \((3h - 5m)(k) = -8k + 16\), Domain: \(\boxed{(-\infty, \infty)}\)