Problem 20 - College Algebra with Corequisite Support

Question

For the following exercises, use each set of functions to find \(f(g(h(x)))\). Simplify your answers. Given \(f(x)=\frac{1}{x}\) and \(g(x)=x-3,\) find the following: (a) \((f \circ g)(x)\) (b) the domain of \((f \circ g)(x)\) in interval notation (c) \((g \circ f)(x)\) (d) the domain of \((g \circ f)(x)\) (e) \(\left(\frac{f}{g}\right)(x)\)

Step-by-Step Solution

Answer

(a) \(\frac{1}{x-3}\); (b) \((-\infty,3)\cup(3,\infty)\); (c) \(\frac{1}{x}-3\); (d) \((-\infty,0)\cup(0,\infty)\); (e) \(\frac{1}{x(x-3)}\).

Step 1: Find \((f \circ g)(x)\)

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). The function \(g(x) = x - 3\) is substituted into \(f(x) = \frac{1}{x}\) to get \(f(g(x)) = \frac{1}{x-3}\).

Step 2: Determine the Domain of \((f \circ g)(x)\)

The function \(f(g(x)) = \frac{1}{x-3}\) is undefined when the denominator is zero. Therefore, solve \(x - 3 = 0\) to find \(x = 3\). Exclude this value from the domain. The domain is all real numbers except \(x = 3\), which is written as \((-\infty, 3) \cup (3, \infty)\) in interval notation.

Step 3: Find \((g \circ f)(x)\)

To find \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). The function \(f(x) = \frac{1}{x}\) is substituted into \(g(x) = x - 3\) to get \(g(f(x)) = \frac{1}{x} - 3\).

Step 4: Determine the Domain of \((g \circ f)(x)\)

The function \(g(f(x)) = \frac{1}{x} - 3\) is undefined when the denominator is zero. Therefore, solve \(x = 0\) and exclude this value from the domain. The domain is all real numbers except \(x = 0\), which is written as \((-\infty, 0) \cup (0, \infty)\) in interval notation.

Step 5: Find \(\left(\frac{f}{g}\right)(x)\)

The function \(\left(\frac{f}{g}\right)(x)\) is given by dividing \(f(x)\) by \(g(x)\). Thus, \(\left(\frac{f}{g}\right)(x) = \frac{1/x}{x - 3}\). Simplify this expression to get \(\frac{1}{x(x - 3)}\).

Key Concepts

Algebraic Functions

Algebraic functions are the building blocks of calculus and mathematical analysis. They involve combinations of constants and variables using arithmetic operations like addition, subtraction, multiplication, division, and taking roots. These functions can be linear or nonlinear and provide a way to model real-world phenomena.
Let's look at the functions provided in the exercise:

• First, we have the function, \(f(x) = \frac{1}{x}\), which is a rational function. It is called so because it can be expressed as the ratio of two polynomials, \(1\) and \(x\).
• Next is \(g(x) = x - 3\), a linear function. This means it forms a straight line when graphed and involves constant change.

Combining functions like \((f \circ g)(x)\) means you plug one function into another. This is known as "function composition." This versatility allows you to create more complex functions and solve various practical problems.

Domain and Range

When dealing with functions, it's crucial to understand the concepts of domain and range. These terms help define the boundaries within which a function operates.
The domain of a function refers to all the possible input values \(x\) can take. For example:

• In \((f \circ g)(x) = \frac{1}{x-3}\), the denominator cannot be zero, which means \(x - 3 eq 0\). This narrows down admissible \(x\) values to all real numbers except \(x = 3\).
• Thus, in interval notation, the domain is represented as \((-\infty, 3) \cup (3, \infty)\).

The range, on the other hand, includes all possible output values after applying the function's formula to its domain. Practicing how to find the domain and range not only helps in understanding function restrictions but also simplifies further calculations.

Simplification of Expressions

Simplifying expressions is central to making complex algebraic problems more manageable. It involves reducing expressions to their simplest form while keeping their value intact.

• Consider the expression \(\left(\frac{f}{g}\right)(x) = \frac{1/x}{x - 3}\). Simplifying this requires understanding and following the rules of fractions and algebraic manipulation.
• Divide \(1/x\) by \(x - 3\) to get \(\frac{1}{x(x - 3)}\). This simplification step helps avoid division errors and miscalculations, making the expression easier to work with.

Try breaking down expressions step by step, looking for common factors, applying the distributive property, or combining like terms. With practice, simplification of expressions becomes an intuitive and essential skill in math and beyond.