Problem 12 - Precalculus with Limits

Question

In Exercises 9-16, find (a) \((f+g)(x)\), (b) \((f-g)(x)\), (c) \((fg)(x)\), and (d) \((f/g)(x). What is the domain of \)f/g\(? \)f(x) = 3x + 1\(, \)g(x) = 5x - 4$

Step-by-Step Solution

Answer

\((f+g)(x) = 8x - 3, \((f-g)(x) = -2x + 5, \((fg)(x) = 15x^2 - 7x - 4, \((f/g)(x) = (3x + 1) / (5x - 4). The domain of f/g is all real numbers except x = 4/5.

Step 1: Addition of functions

The sum of the functions \(f\) and \(g\) is given by: \[(f+g)(x) = f(x) + g(x) = (3x + 1) + (5x - 4) = 8x - 3.\]

Step 2: Subtraction of functions

The difference of the functions \(f\) and \(g\) is given by: \[(f-g)(x) = f(x) - g(x) = (3x + 1) - (5x - 4) = -2x + 5.\]

Step 3: Multiplication of functions

The product of the functions \(f\) and \(g\) is given by: \[(fg)(x) = f(x) * g(x) = (3x + 1) * (5x - 4) = 15x^2 - 7x - 4.\]

Step 4: Division of functions

The quotient of the functions \(f\) and \(g\) is given by: \[(f/g)(x) = f(x) / g(x) = (3x + 1) / (5x - 4).\] Note that this function is not defined when \(g(x) = 0\), which happens when \(x = 4/5\).

Step 5: Determine the domain of f/g

The domain of the function \(f/g\) will be all real numbers except for the value that makes \(g(x) = 0\), which is \(x = 4/5\). So the domain of \(f/g\) is \(-\infty, 4/5) \cup (4/5, +\infty\).

Key Concepts

Composite Functions

Understanding composite functions is a fundamental aspect of precalculus, essential for higher mathematics. A composite function is created when one function is applied to the results of another function.

For instance, if we have two functions, \(f\) and \(g\), and we want to compute \(f(g(x))\), we are effectively finding the value of \(g\) at \(x\), and then applying \(f\) to that result. Think of it as a two-step process: whatever \(g\) spits out, \(f\) immediately takes in as its input.

When working with composite functions, one must be mindful of the domain of each function to ensure the composite is well-defined. The order of function application is critical here, as \(f(g(x))\) may not be the same as \(g(f(x))\). It's like following a recipe; the sequence of steps can dramatically affect the final dish.

Domain of a Function

The domain of a function indicates the set of all possible input values (commonly referred to as \(x\) values) that the function can accept without problems. Determining the domain is like ensuring you don't invite a friend with a peanut allergy to a peanut festival; you want to identify which \(x\)-values don't cause trouble.

In our example from the exercise, for \(f/g\), the domain excludes \(x\) values that make \(g(x) = 0\), because dividing by zero is undefined. In simple terms, if you have a function that divides by another, like \(f/g\), you must exclude any \(x\)-values from the domain that turn the denominator \(g(x)\) into zero. For our specific functions, \(g(x)\) becomes zero at \(x = 4/5\), so the domain of \(f/g\) is all real numbers except \(4/5\).

Remember, the domain is like the guest list for a function's party; we carefully select who gets in to avoid disasters.

Function Arithmetic

Similar to traditional arithmetic with numbers, function arithmetic involves operations like addition, subtraction, multiplication, and division with functions. It's comparable to combining different ingredients in cooking to create new flavors.

When functions \(f\) and \(g\) are added, \(f+g\), we simply add their outputs for the same input \(x\). Subtraction, \(f-g\), follows the same principle but instead subtracts the outputs. Multiplication, \(fg\), finds the product of the outputs, and division, \(f/g\), finds the quotient, provided the divisor function \(g\) is not zero at that particular \(x\).

Using the given exercise, the sum \(f+g\) gives us a straightforward linear function, while the product \(fg\) leads to a quadratic. Different operations lead to functions with various characteristics and complexities, illustrating the beauty and symmetry of function arithmetic.