Problem 23 - A Graphical Approach to Precalculus with Limits

Question

For each pair of fimetions, (a) find ( \(f+g)(x),(f-g)(x),\) and \((f g)(x) ;\) (b) give the domains of the functions in part (a); (c) find \(\frac{t}{x}\) and give its domain; (d) find \(f \circ g\) and give is domain: and (e) find \(g \circ f\) and give its domain. Do not use a calculator. $$f(x)=4 x-1, g(x)=6 x+3$$

Step-by-Step Solution

Answer

Functions: \((f+g)(x) = 10x + 2\), \((f-g)(x) = -2x - 4\), \((fg)(x) = 24x^2 + 6x - 3\). Domain: \(\mathbb{R}\), except \(\mathbb{R} \setminus \{-1/2\}\) for \(\frac{f}{g}(x)\). Compositions: \(f \circ g = 24x + 11\), \(g \circ f = 24x - 3\), both with domain \(\mathbb{R}\).

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

To find \((f+g)(x)\), add the two functions:\[(f+g)(x) = f(x) + g(x) = (4x - 1) + (6x + 3)\]Now simplify the equation:\[(f+g)(x) = 4x + 6x - 1 + 3 = 10x + 2\]The resulting function is \((f+g)(x) = 10x + 2\).

Step 2: Find (f-g)(x)

Subtract \(g(x)\) from \(f(x)\):\[(f-g)(x) = f(x) - g(x) = (4x - 1) - (6x + 3)\]Now simplify the equation:\[(f-g)(x) = 4x - 6x - 1 - 3 = -2x - 4\]The resulting function is \((f-g)(x) = -2x - 4\).

Step 3: Find (fg)(x)

To find \((fg)(x)\), multiply the two functions:\[(fg)(x) = f(x) \, g(x) = (4x - 1)(6x + 3)\]Now, expand and simplify the product:\[(fg)(x) = 4x(6x) + 4x(3) - 1(6x) - 1(3)\]\[= 24x^2 + 12x - 6x - 3\]\[= 24x^2 + 6x - 3\]The resulting function is \((fg)(x) = 24x^2 + 6x - 3\).

Step 4: Determine Domains for Part (a)

The domain for \(f(x)\), \(g(x)\), and the resulting functions in part (a), \((f+g)(x)\), \((f-g)(x)\), and \((fg)(x)\) are all the set of real numbers, \(\mathbb{R}\), because there are no restrictions such as divisions by zero or square roots of negative numbers.

Step 5: Find \(\frac{f}{g}(x)\) and its Domain

Calculate the function:\[\frac{f}{g}(x) = \frac{4x - 1}{6x + 3}\]The domain of \(\frac{f}{g}(x)\) consists of all real numbers except where the denominator is zero:\[6x + 3 = 0\]Solve for \(x\):\[6x = -3 \, \Rightarrow \, x = -\frac{1}{2}\]Therefore, the domain is \(\mathbb{R} \setminus \{-\frac{1}{2}\}\).

Step 6: Find f ∘ g(x) and its Domain

The composition \(f \circ g\) means \(f(g(x))\):\[f(g(x)) = f(6x + 3) = 4(6x + 3) - 1\]Simplify:\[= 24x + 12 - 1 = 24x + 11\]The function \(g(x)\) has no restrictions, so the domain of \(f \circ g(x)\) is \(\mathbb{R}\).

Step 7: Find g ∘ f(x) and its Domain

The composition \(g \circ f\) means \(g(f(x))\):\[g(f(x)) = g(4x - 1) = 6(4x - 1) + 3\]Simplify:\[= 24x - 6 + 3 = 24x - 3\]The function \(f(x)\) has no restrictions, so the domain of \(g \circ f(x)\) is \(\mathbb{R}\).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values (usually denoted as \(x\)) for which the function is defined. When dealing with functions like polynomials, you generally find the domain to be all real numbers, \(\mathbb{R}\), unless there are places where the function is not defined, such as divisions by zero or square roots of negative numbers.

For instance, in the original exercise, the function \(\frac{f}{g}(x) = \frac{4x - 1}{6x + 3}\) must be examined carefully because of its denominator. Here, you need to ensure the denominator \(6x + 3eq0\), which makes \(x eq -\frac{1}{2}\). Therefore, the domain is all real numbers except \(-\frac{1}{2}\).

In general, when managing functions, always look out for instances where denominators become zero. Similarly, be cautious with any operations that could involve the square roots of negative numbers, as this may further restrict the domain.

Function Composition

Function composition is a process that involves combining two functions to create a new function. In notation, it is represented as \(f \circ g(x)\), meaning you apply \(g\) first, then take the output of \(g\) and input it into \(f\).

To successfully execute function composition, understanding both the inner and outer functions is necessary. In the example with \(f(x) = 4x - 1\) and \(g(x) = 6x + 3\), composing \(f \circ g\) involves substituting \(g(x)\) into \(f(x)\):

• Compute \(g(x) = 6x + 3\)
• Substitute \(g(x)\) into \(f(x)\) to get \(f(g(x)) = 4(6x + 3) - 1\)
• Simplify to find \(f(g(x)) = 24x + 11\)

It’s crucial here to ensure both functions can operate within each other's domains, which typically remains the set of all real numbers for polynomial functions unless otherwise restricted.

Remember that the domain of the composed function stems from the domain of the inside function \((g)\) and any new restrictions introduced by the outer function \((f)\). In our solution, since \(g(x)\) is a linear polynomial and unrestricted, \(f \circ g(x)\) remains defined across \(\mathbb{R}\).

Polynomial Functions

Polynomial functions are among the simplest yet most important types of functions. They are defined as expressions of the form \(anx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\), where \(n\) is a non-negative integer and \(a_n, a_{n-1}, \ldots, a_1, a_0\) are constants with \(a_n eq 0\).

In the exercise, functions \(f(x) = 4x - 1\) and \(g(x) = 6x + 3\) are both simple linear polynomials of the form \(ax + b\). They involve straightforward operations:

• Addition: \((f+g)(x) = 10x + 2\)
• Subtraction: \((f-g)(x) = -2x - 4\)
• Multiplication: \((fg)(x) = 24x^2 + 6x - 3\)

Working with them often means combining like terms and carrying out basic arithmetic directly on the coefficients.

Polynomials are nice to work with because their domains are vast; unless you introduce fractions whereby a variable appears in the denominator, signals for possible restrictions are minimal. The behavior and domain properties of polynomials considerably simplify many analyses and operations students encounter in algebra and calculus.