Problem 7 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 3-8: Find (a) \((f+g)(x),(f-g)(x),(f g)(x)\), and \((f / g)(x)\) (b) the domain of \(f+g, f-g\), and \(f g\) (c) the domain of \(f / g\) $$ f(x)=\frac{2 x}{x-4}, \quad g(x)=\frac{x}{x+5} $$

Step-by-Step Solution

Answer

(f+g)(x) = \frac{3x^2 + 6x}{(x-4)(x+5)}, (f-g)(x) = \frac{x^2 + 14x}{(x-4)(x+5)}, fg(x) = \frac{2x^2}{(x-4)(x+5)}, \frac{f}{g}(x) = \frac{2(x+5)}{x-4}; Domain of (f+g), (f-g), fg: (-\infty, -5) \cup (-5,4) \cup (4,\infty); Domain of \frac{f}{g}: (-\infty,-5) \cup (-5,0) \cup (0,4) \cup (4,\infty).

Step 1: Add the Functions

To find \(f+g\)(x), we add the functions \ f(x)\ and \ g(x)\:\[ (f+g)(x) = \frac{2x}{x-4} + \frac{x}{x+5} \]Get a common denominator, which is \ (x-4)(x+5)\:\[ (f+g)(x) = \frac{2x(x+5) + x(x-4)}{(x-4)(x+5)} \]Simplify to:\[ (f+g)(x) = \frac{2x^2 + 10x + x^2 - 4x}{(x-4)(x+5)} = \frac{3x^2 + 6x}{(x-4)(x+5)} \].

Step 2: Subtract the Functions

To find \(f-g\)(x), subtract \ g(x) \ from \ f(x)\:\[ (f-g)(x) = \frac{2x}{x-4} - \frac{x}{x+5} \]Use a common denominator \(x-4)(x+5)\):\[ (f-g)(x) = \frac{2x(x+5) - x(x-4)}{(x-4)(x+5)} \]Simplify to:\[ (f-g)(x) = \frac{2x^2 + 10x - x^2 + 4x}{(x-4)(x+5)} = \frac{x^2 + 14x}{(x-4)(x+5)} \].

Step 3: Multiply the Functions

To find \ (f \, g)(x) \, multiply \ f(x) \ and \ g(x)\:\[ (f \, g)(x) = \frac{2x}{x-4} \cdot \frac{x}{x+5} \]Multiply the numerators and the denominators:\[ (f \, g)(x) = \frac{2x^2}{(x-4)(x+5)} \].

Step 4: Divide the Functions

To find \(\frac{f}{g}\)(x), divide \ f(x) \ by \ g(x)\:\[ \left(\frac{f}{g}\right)(x) = \frac{\frac{2x}{x-4}}{\frac{x}{x+5}} \]Multiply by the reciprocal:\[ \left(\frac{f}{g}\right)(x) = \frac{2x}{x-4} \cdot \frac{x+5}{x} \]Simplify to:\[ \left(\frac{f}{g}\right)(x) = \frac{2(x+5)}{x-4} \].

Step 5: Determine the Domain of Sum, Difference, and Product

For \ f+g, f-g, \ and \ f g \, the domain is all the x-values for which both \ f(x) \ and \ g(x) \ are defined.The domain excludes points where any denominator is zero:\[ x eq 4 \quad \text{and} \quad x eq -5 \].Thus, the domain is \ (-\infty, -5) \cup (-5,4) \cup (4,\infty) \.

Step 6: Determine the Domain of the Quotient

The domain of \ \frac{f}{g} \ includes all values where \ f(x) \ and \ g(x) \ are defined and \ g(x) \ is non-zero.The domain excludes points where any denominator is zero and where \ g(x) = 0\.\[ g(x) = \frac{x}{x+5} = 0 \Rightarrow x = 0 \].Thus, the domain is \ (-\infty,-5) \cup (-5,0) \cup (0,4) \cup (4,\infty) \.

Key Concepts

Domain of Functions

The domain of a function refers to all the possible input values (typically represented as \(x\)) that will produce valid outputs for a given function. Think of it as the set of "permitted" values for \(x\). For functions involving fractions, the denominator cannot be zero, as division by zero is undefined.

When determining the domain of functions, especially those involving rational expressions, it's crucial to identify the values that make any denominator zero and exclude them from the domain.

For instance, if we have \(f(x) = \frac{2x}{x-4}\) and \(g(x) = \frac{x}{x+5}\), the denominators become zero at \(x-4=0\) and \(x+5=0\) respectively, leading to \(x = 4\) and \(x = -5\). Consequently, these values are not included in the domain. For rational operations between \(f(x)\) and \(g(x)\), like addition or multiplication, we need to find the common domain of both functions, ensuring that neither denominator evaluates to zero.

Addition of Functions

When adding two functions, \(f(x) + g(x)\), we combine their outputs at the same input values. This operation involves finding a common denominator if the functions are rational expressions.

Consider the functions \(f(x) = \frac{2x}{x-4}\) and \(g(x) = \frac{x}{x+5}\). To perform the addition, we obtain:

• A common denominator: \((x-4)(x+5)\)
• Rewrite both functions with this common denominator.
  Combine and simplify the numerators

This process results in another rational function: \[ (f+g)(x) = \frac{3x^2 + 6x}{(x-4)(x+5)} \] Remember, the domain of \(f+g\) excludes \(x = 4\) and \(x = -5\), where the original denominators vanish.

Subtraction of Functions

Subtracting functions, \(f(x) - g(x)\), means taking the outcome of \(f(x)\) and subtracting \(g(x)\) evaluated at the same input \(x\). Similar to addition, subtraction requires a common denominator when working with fractions.

For \(f(x) = \frac{2x}{x-4}\) and \(g(x) = \frac{x}{x+5}\), we perform subtraction as follows:

• Obtain a common denominator, \((x-4)(x+5)\)
• Convert each function with this denominator and subtract the numerators Simplify the final expression

This gives us: \[ (f-g)(x) = \frac{x^2 + 14x}{(x-4)(x+5)} \] The domain of \(f-g\) also excludes \(x = 4\) and \(x = -5\), adhering to the restrictions from each function's denominator.

Multiplication of Functions

In function multiplication, the outputs of the functions are multiplied for the same input. For two rational functions, the multiplication process is straightforward since we multiply numerators and denominators directly.

Using \(f(x) = \frac{2x}{x-4}\) and \(g(x) = \frac{x}{x+5}\), the multiplication is done as:

• Multiply the numerators: \(2x \cdot x = 2x^2\)
• Multiply the denominators: \((x-4)(x+5)\)

This results in: \[ (f \cdot g)(x) = \frac{2x^2}{(x-4)(x+5)} \] The domain must exclude \(x = 4\) and \(x = -5\) because they make either denominator undefined.

Division of Functions

Dividing functions is similar to multiplying, but instead of direct multiplication, we multiply by the reciprocal. Division between two functions, \(\frac{f(x)}{g(x)}\) involves certain extra considerations since the denominator \(g(x)\) should also not be zero.

For \(f(x) = \frac{2x}{x-4}\) and \(g(x) = \frac{x}{x+5}\), division involves:

• Inverting \(g(x)\) for the reciprocal: \(\frac{x+5}{x}\)
• Multiply \(f(x)\) by the reciprocal of \(g(x)\)
• Simplify: \[ \frac{2x}{x-4} \cdot \frac{x+5}{x} = \frac{2(x+5)}{x-4} \]

The domain of \(\frac{f}{g}\) is more restrictive. Besides \(x = 4\) and \(x = -5\), it also cannot be \(x = 0\), as this would make \(g(x) = 0\) leading to division by zero. Hence, these values are excluded from the domain.