Problem 36 - Precalculus Essentials

Question

Find \(f+g, f-g,\) fg, and \(\frac{f}{x}\). Determine the domain for each function. $$f(x)=6 x^{2}-x-1, g(x)=x-1$$

Step-by-Step Solution

Answer

The functions resulting from the operations are \( (f+g)(x) = 6 x^{2}-2, (f-g)(x) = 6 x^{2} - 2x, fg(x) = 6x^{3} - 7x^{2} + 2x - 1, \frac{f}{x} = 6x - 1 - \frac{1}{x}\). The domains for \(f+g\), \(f-g\) and \(fg\) are \(-\infty,+\infty\), for \( \frac{f}{x}\) it is \(-\infty,0) \cup (0,+\infty\).

Step 1: Addition of the functions

Adding the two functions \(f(x)\) and \(g(x)\) together involves adding the matching components of each function. This results in \( (f+g)(x) = 6 x^{2}-x-1 + x-1 = 6 x^{2}-2.\)

Step 2: Subtraction of the functions

Subtracting the function \(g(x)\) from \(f(x)\) involves subtracting the matching components of each function. This results in \( (f-g)(x) = 6 x^{2}-x-1 - (x-1) = 6 x^{2}-2x.\)

Step 3: Multiplication of the functions

Multiplication of the functions \(f(x)\) and \(g(x)\) involves multiplying the two entire functions together. This will result in \( fg(x) = (6 x^{2}-x-1)(x-1) = 6x^{3} - 7x^{2} + 2x - 1. \)

Step 4: Division of the functions

Division of the functions \( f(x) \) by \( x \) involves dividing each component of \( f(x) \) by \( x \). This will result in \( \frac{f}{x} = \frac{6 x^{2}-x-1}{x} = 6x - 1 - \frac{1}{x} \)

Step 5: Determine the domain for each function

The domain for each function is determined by the values of \(x\) for which the function is defined. All polynomials are defined for all real numbers, so the domains of \(f+g\), \(f-g\) and \(fg\) are all \(-\infty,+\infty\). The function \( \frac{f}{x} \) is undefined if \(x=0\), so its domain is \(-\infty,0) \cup (0,+\infty\).

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. In simpler terms, they are expressions like \(6x^2 - x - 1\) or \(x - 1\). These are examples of polynomials with one variable \(x\).

• The degree of a polynomial is determined by its highest exponent. For instance, in \(f(x) = 6x^2 - x - 1\), the polynomial's degree is 2 because the highest power of \(x\) is 2.
• Polynomials can have various terms and the number of terms defines their type: monomial (one term), binomial (two terms), or trinomial (three terms).

Polynomials are fundamental in algebra and calculus because they can easily model a wide array of natural phenomena and are generally simple to manipulate with addition, subtraction, multiplication, and division.

Function Addition

Function addition is one of the basic operations you can perform on two or more functions. When adding functions, you simply add the values of the functions for each value of \(x\).
If you have two functions, \(f(x)\) and \(g(x)\), their sum \((f + g)(x)\) is calculated by adding their expressions:

• \(f(x) = 6x^2 - x - 1\)
• \(g(x) = x - 1\)
• The addition result is \((f+g)(x) = 6x^2 - x - 1 + x - 1 = 6x^2 - 2\).

Function addition combines the like terms (the coefficients of the same powers of \(x\)) to simplify the expression.

Function Subtraction

Function subtraction works much like function addition, but instead of adding the values, you subtract them. This operation can show differences in how two functions behave across their domains.
For functions \(f(x)\) and \(g(x)\), you calculate the difference \((f - g)(x)\):

• \(f(x) = 6x^2 - x - 1\)
• \(g(x) = x - 1\)
• The subtraction result is \((f-g)(x) = 6x^2 - x - 1 - (x - 1) = 6x^2 - 2x\).

Subtracting \(g(x)\) from \(f(x)\) involves reversing the sign of \(g(x)\) first before combining like terms.

Function Multiplication

Multiplying functions involves taking two functions and multiplying their expressions together. This creates a product function that represents a different set of values across the shared domain of the two functions.
With the given functions \(f(x)\) and \(g(x)\), the multiplication \(fg(x)\) is done as follows:

• \(f(x) = 6x^2 - x - 1\)
• \(g(x) = x - 1\)
• The product result is \((fg)(x) = (6x^2 - x - 1)(x - 1)= 6x^3 - 7x^2 + 2x - 1\).

Each term of one function multiplies every term of the other, and like terms are combined as part of simplification.

Function Division

Function division is slightly more complex, especially when you divide by a function that can equal zero, which influences the domain. Division involves dividing one function, like \(f(x)\), by another, like \(g(x)\).
For the problem given:

• The division of \(f(x)\) by \(x\) is expressed as: \(\frac{f}{x} = \frac{6x^2 - x - 1}{x} = 6x - 1 - \frac{1}{x}\).

Notice the \(x\) in the denominator? This means the value 0 cannot be in the domain, as it would make the function undefined at that point.

Domain of a Function

The domain of a function is the set of all possible input values (typically \(x\)) that the function can accept without leading to any mathematical issues like division by zero or taking the square root of a negative number.

• For polynomial functions like \(f+g\), \(f-g\), and \(fg\), the domain tends to be all real numbers, or \((-\infty, +\infty)\), because they are defined for every real number.
• However, division functions like \(\frac{f}{x}\) have restricted domains because division by zero is undefined. Therefore, the domain excludes \(x = 0\), resulting in \((-\infty, 0) \cup (0, +\infty)\).

This understanding of domain helps to ensure that functions you work with remain valid mathematic expressions.