Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 11-14: If \(a\) is a positive real number, find (a) \(g\left(\frac{1}{a}\right)\) (b) \(\frac{1}{g(a)}\) (c) \(g(\sqrt{a})\) (d) \(\sqrt{g(a)}\) $$ g(x)=\frac{2 x}{x^{2}+1} $$

Step-by-Step Solution

Answer

(a) \( \frac{2a}{1+a^2} \), (b) \( \frac{a^2+1}{2a} \), (c) \( \frac{2\sqrt{a}}{a+1} \), (d) \( \sqrt{\frac{2a}{a^2+1}} \)."

Step 1: Understand the Function

The given function is \( g(x) = \frac{2x}{x^2 + 1} \). We'll use this definition to solve each part (a), (b), (c), and (d) by substituting the given expressions into this function.

Step 2: Solve Part (a) - Find \( g\left(\frac{1}{a}\right) \)

To find \( g\left(\frac{1}{a}\right) \), substitute \( \frac{1}{a} \) into the function:\[g\left(\frac{1}{a}\right) = \frac{2\left(\frac{1}{a}\right)}{\left(\frac{1}{a}\right)^2 + 1} = \frac{\frac{2}{a}}{\frac{1}{a^2} + 1}.\]Simplifying, we get:\[g\left(\frac{1}{a}\right) = \frac{\frac{2}{a}}{\frac{1 + a^2}{a^2}} = \frac{2a}{1 + a^2}.\]

Step 3: Solve Part (b) - Find \( \frac{1}{g(a)} \)

First find \( g(a) \) using the function:\[g(a) = \frac{2a}{a^2 + 1}.\]Now find the reciprocal:\[\frac{1}{g(a)} = \frac{1}{\left(\frac{2a}{a^2 + 1}\right)} = \frac{a^2 + 1}{2a}.\]

Step 4: Solve Part (c) - Find \( g(\sqrt{a}) \)

Substitute \( \sqrt{a} \) into the function:\[g(\sqrt{a}) = \frac{2\sqrt{a}}{(\sqrt{a})^2 + 1} = \frac{2\sqrt{a}}{a + 1}.\]

Step 5: Solve Part (d) - Find \( \sqrt{g(a)} \)

First find \( g(a) \) as done previously:\[g(a) = \frac{2a}{a^2 + 1}.\]Then find the square root:\[\sqrt{g(a)} = \sqrt{\frac{2a}{a^2 + 1}}.\]

Key Concepts

Understanding Reciprocal Functions

Reciprocal functions are unique in algebra because they involve flipping or inverting the input value. When we talk about the reciprocal of a number or expression, we're simply referring to the inverse of that number. For example, the reciprocal of 2 is \( \frac{1}{2} \).
In reciprocal functions, if you have a function like \( y = \frac{f(x)}{g(x)} \), finding the reciprocal function involves inverting this relationship. For example, if \( g(x) = \frac{2x}{x^2 + 1} \), then the reciprocal function \( \frac{1}{g(x)} \) would be simply \( \frac{x^2 + 1}{2x} \).
Thus, understanding how to switch the roles of the numerator and denominator is key when working with reciprocals.

Function Substitution Techniques

Function substitution is a powerful tool in algebra that allows you to plug in different expressions or values into a function. This technique is essentially about replacing variables within a function.
Imagine our function \( g(x) = \frac{2x}{x^2 + 1} \), and we needed to compute \( g\left(\frac{1}{a}\right) \). The process involves replacing every \( x \) in the function with \( \frac{1}{a} \).

• First, substitute \( x = \frac{1}{a} \) into the function: \( \frac{2(\frac{1}{a})}{(\frac{1}{a})^2 + 1} \).
• Simplify step by step to get the final expression: \( \frac{2a}{1 + a^2} \).

This approach allows you to explore how functions behave with different arguments or inputs.

The Process of Function Simplification

Function simplification is all about making complex expressions easier to understand and work with. When dealing with algebraic functions, simplification ensures that we find an expression in its simplest form.
Suppose we are dealing with the function \( g(x) = \frac{2x}{x^2 + 1} \) and need to simplify the expression for various values. For example, let’s simplify \( g\left(\frac{1}{a}\right) \). After substitution, the expression initially reads as \( \frac{\frac{2}{a}}{\frac{1}{a^2} + 1} \).

• Find a common denominator: Transform \( \frac{1}{a^2} + 1 \) into \( \frac{1 + a^2}{a^2} \).
• Multiply through by \( a^2 \) to eliminate the fractions within fractions.
• Achieve a simplified form: \( \frac{2a}{1 + a^2} \).

The key is to methodically work through the steps, simplifying at each stage.

Exploring Real Numbers in Functions

Real numbers are the foundation of mathematics that include both rational and irrational numbers. In the context of functions, understanding real numbers is crucial as it affects the domain and range we work within.
A real number can be any number you can find on the number line, including integers, fractions, and non-repeating decimals. When we describe our function \( g(x) = \frac{2x}{x^2 + 1} \) in the domain of real numbers, we're specifying the set of inputs \( x \) can take.

• All positive real numbers work for the given exercise since \( a \) is defined as positive.
• In cases involving real numbers, ensuring the function remains defined (like ensuring the denominator isn’t zero) is important.

As you engage with functions, real numbers expand the possibilities of solutions and make algebraic operations extensive and practical.