Question

Write \(p(x)=1 / \sqrt{x^{2}+1}\) as a composite of three functions in two different ways.

Short answer

1st way: \( f(u) = \frac{1}{u}, g(v) = \sqrt{v}, h(x) = x^2 + 1 \). 2nd way: \( f(v) = \frac{1}{\sqrt{v}}, g(u) = u + 1, h(x) = x^2 \).

Step-by-step solution

Identifying the Outermost Function

The function given is \( p(x) = \frac{1}{\sqrt{x^2 + 1}} \). The outermost operation applied is taking the reciprocal. Therefore, identify the outer function as \( f(u) = \frac{1}{u} \).

Recognizing the Intermediate Function

Next, identify the second operation as taking the square root, suggesting the intermediate function \( g(v) = \sqrt{v} \).

Determining the Innermost Function

The innermost operation is \( x^2 + 1 \). Thus, the innermost function is \( h(x) = x^2 + 1 \).

Writing the Composite in the First Way

Combine the functions identified in Steps 1-3: \( p(x) = f(g(h(x))) \), where \( f(u) = \frac{1}{u} \), \( g(v) = \sqrt{v} \), and \( h(x) = x^2 + 1 \).

Identifying the Second Way's Innermost Function

Consider \( h(x) = x^2 \) as the innermost function, moving the constant operation outside.

Recognizing the Second Way's Intermediate Function

The second intermediate function includes the addition, represented as \( g(u) = u + 1 \).

Updating the Second Way's Outer Function

Now, the outermost function will handle both the square root and reciprocal operations: \( f(v) = \frac{1}{\sqrt{v}} \).

Writing the Composite in the Second Way

Combine these functions for the second way: \( p(x) = f(g(h(x))) \), where \( h(x) = x^2 \), \( g(u) = u + 1 \), and \( f(v) = \frac{1}{\sqrt{v}} \).

Key concepts

Function Composition

Function composition is a powerful concept in mathematics that allows you to combine two or more functions into a single expression. By understanding this, we can simplify complex problems and break them down into manageable parts. Composition involves using the output of one function as the input of another.

In the given exercise, function composition is used to express a function, \( p(x) \), in more than one way. Each method uses different composite functions: outer, intermediate, and inner. For the first composition:

• The outer function is \( f(u) = \frac{1}{u} \), which signifies taking the reciprocal of \( u \).
• The intermediate function \( g(v) = \sqrt{v} \) involves taking the square root.
• The innermost function \( h(x) = x^2 + 1 \) involves a polynomial expression.

The composition creates \( p(x) = f(g(h(x))) \). Each function acts in a sequence, providing a step-by-step transformation from the innermost to the outermost operation.

Mathematical Functions

Mathematical functions are the backbone of algebra and calculus. They describe the relationship between different variables by associating every input with exactly one output. Functions can be simple, like linear and quadratic functions, or they can involve more complex operations such as polynomial or trigonometric functions. Understanding how to identify and work with these functions is crucial for solving problems efficiently.

In our exercise, three specific types of mathematical functions are involved. The innermost function \( h(x) = x^2 + 1 \) is a polynomial function. Polynomials are expressions made of variables and constants combined using addition, subtraction, and multiplication. Next, the intermediate function \( g(v) = \sqrt{v} \) is a radical function since it involves a square root. Lastly, the outer function \( f(u) = \frac{1}{u} \) represents a rational function, highlighting the concept of reciprocals. By learning these functions, students can simplify expressions and calculate values step-by-step.

Reciprocal and Square Root

The reciprocal and square root operations are essential concepts in algebra. Understanding these operations helps in manipulating expressions, solving equations, and understanding the characteristics of functions.

The reciprocal of a number is simply \( 1 \) divided by that number (\( \frac{1}{a}\)). Reciprocals are useful in simplifying expressions and solving equations. In the exercise, the function \( f(u) = \frac{1}{u} \) demonstrates this operation.

On the other hand, square roots involve finding a number which, when multiplied by itself, gives the original number (\( \sqrt{a} \)). In the problem, \( g(v) = \sqrt{v} \) relates to this operation. Recognizing the difference between each operation is crucial, as each serves a unique purpose in functional composition and can significantly alter the overall expression.

• Reciprocals simplify expressions dramatically by inverting values.
• Square roots transform expressions by scaling them down, often simplifying the roots of polynomial expressions.

Understanding both operations allows you to break down and analyze complex functions easily.