Problem 12 - Precalculus

Question

In Exercises \(11-20,\) use the pair of functions \(f\) and \(g\) to find the domain of the indicated function then find and simplify an expression for it. -\((f+g)(x)\) -\((f-g)(x)\) -\((f g)(x)\) -\(\left(\frac{f}{g}\right)(x)\) $$f(x)=1-4 x \text { and } g(x)=2 x-1$$

Step-by-Step Solution

Answer

Domains: (f+g)(x), (f-g)(x), and (fg)(x) are all reals; (f/g)(x) domain is all reals except x = 1/2. Expressions: (f+g)(x) = -2x, (f-g)(x) = -6x + 2, (fg)(x) = -8x^2 + 6x - 1, (f/g)(x) = (1-4x)/(2x-1).

Step 1: Understanding Domain for (f+g)(x)

The domain of the function \((f+g)(x) = (1-4x) + (2x-1)\) is the intersection of the domains of \(f\) and \(g\). Both \(f(x)\) and \(g(x)\) are linear functions, which have a domain of all real numbers \((-\infty, \infty)\). So the domain for \((f+g)(x)\) is also all real numbers.

Step 2: Simplifying Expression for (f+g)(x)

Simplify the expression by combining like terms: \((f+g)(x) = 1 - 4x + 2x - 1 = -2x\). Therefore, \((f+g)(x) = -2x\).

Step 3: Finding Domain of (f-g)(x)

Just as with addition, the domain for \((f-g)(x)\) is found by considering the intersection of the domains of \(f\) and \(g\). Since both functions are linear, the domain is all real numbers \((-\infty, \infty)\).

Step 4: Simplifying Expression for (f-g)(x)

Subtract \(g(x)\) from \(f(x)\): \((f-g)(x) = (1-4x) - (2x-1) = 1 - 4x - 2x + 1 = -6x + 2\). Thus, \((f-g)(x) = -6x + 2\).

Step 5: Domain for (fg)(x)

The domain for \((fg)(x)\) is again the intersection of the domains of \(f\) and \(g\), which is all real numbers \((-\infty, \infty)\) since both functions can take any real number.

Step 6: Simplifying Expression for (fg)(x)

Multiply the expressions for \(f(x)\) and \(g(x)\): \((fg)(x) = (1-4x)(2x-1) = 2x - 1 - 8x^2 + 4x = -8x^2 + 6x - 1\). Thus, \((fg)(x) = -8x^2 + 6x - 1\).

Step 7: Domain for (f/g)(x)

The domain for \(\left(\frac{f}{g}\right)(x)\) includes all real numbers except where \(g(x) = 0\) because division by zero is undefined. Set \(g(x) = 2x - 1 = 0\), solving for \(x\) gives \(x = \frac{1}{2}\). Thus, the domain is all real numbers except \(x = \frac{1}{2}\), or \((-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)\).

Step 8: Simplifying Expression for (f/g)(x)

Divide \(f(x)\) by \(g(x)\) to get \(\left(\frac{f}{g}\right)(x) = \frac{1 - 4x}{2x - 1}\). This is the expression simplified for \(\left(\frac{f}{g}\right)(x)\).

Key Concepts

Linear Functions

Linear functions are one of the simplest forms of functions in mathematics. They have the general equation form of \(f(x) = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
Linear functions are characterized by their straight-line graphs. Importantly, they have a domain of all real numbers, which we write as \((-, )\).
This means that you can input any real number into the linear function, and it will produce a valid output. The simplicity of their domain is helpful when combining linear functions, as any operations (addition, subtraction, multiplication) will also have a domain of all real numbers.
Knowing this makes it straightforward to calculate and understand function operations involving linear functions.

Function Operations

In mathematics, function operations involve combining different functions through addition, subtraction, multiplication, or division. When performing these operations, it is crucial to consider the domain of the resulting function.
- **Addition and Subtraction**: For functions \(f(x)\) and \(g(x)\), their sum \((f+g)(x)\) and difference \((f-g)(x)\) are found by adding or subtracting the function expressions, respectively. The domain of \((f+g)(x)\) and \((f-g)(x)\) is the intersection of the domains of \(f\) and \(g\).
- **Multiplication**: The product \((fg)(x)\) is obtained by multiplying the expressions for \(f(x)\) and \(g(x)\). Like addition and subtraction, the domain is determined by the common values that \(f\) and \(g\) can accept.
- **Division**: The quotient \((f/g)(x)\) is defined as \(\frac{f(x)}{g(x)}\), with an important condition: \(g(x)\) cannot be zero, as division by zero is undefined. Therefore, the domain of \((f/g)(x)\) excludes any values of \(x\) that make \(g(x) = 0\).
Understanding these operations allows us to manipulate complex expressions and determine their domains more easily.

Rational Functions

Rational functions are ratios of two polynomials. Generally, they take the form \(\frac{p(x)}{q(x)}\), where \(p(x)\) and \(q(x)\) are polynomials, and \(q(x) eq 0\).
Analyzing the domain of a rational function is crucial because the function is undefined wherever \(q(x)\) (the denominator) equals zero.
To find these undefined points, we set \(q(x) = 0\) and solve for \(x\). For the exercise given, we found that for \(g(x) = 2x - 1\), the function is undefined at \(x = \frac{1}{2}\). Thus, the domain of \(\frac{f(x)}{g(x)}\) is all real numbers except \(x = \frac{1}{2}\). This breaks down as:

• Values of \(x\) where \(q(x) eq 0\) make the function valid.
• Any \(x\) that satisfies \(q(x) = 0\) must be excluded from the domain.

Rational functions can have complex graphs with vertical asymptotes where the function values approach infinity or negative infinity. By understanding their domain, we can appropriately sketch and interpret these graphs.