Problem 9 - Calculus Single Variable

Question

Let \(F(x)=x^{2}+5, G(x)=(x+1) /(x-1),\) and \(H(x)=2 x-5 .\) Calculate the value of the given function at \(x\). \(G \circ(1 / G)\)

Step-by-Step Solution

Answer

The composition \(G\left(\frac{1}{G(x)}\right)\) simplifies to \(-x\).

Step 1: Define the Functions

We are given the functions as follows: \[ F(x) = x^2 + 5, \quad G(x) = \frac{x+1}{x-1}, \quad H(x) = 2x - 5 \] We need to evaluate \( G \circ \left( \frac{1}{G} \right) \), which is the composition of \( G(x) \) with \( \frac{1}{G(x)} \).

Step 2: Find \(\frac{1}{G(x)}\)

To find \( \frac{1}{G(x)} \), we take the reciprocal of \( G(x) \).Given \( G(x) = \frac{x+1}{x-1} \), the reciprocal is:\[ \frac{1}{G(x)} = \frac{x-1}{x+1} \]

Step 3: Evaluate the Composition \(G\left(\frac{1}{G(x)}\right)\)

Now replace \(x\) in \(G(x)\) with \(\frac{x-1}{x+1}\):\[ G\left(\frac{x-1}{x+1}\right) = \frac{\left(\frac{x-1}{x+1}\right) + 1}{\left(\frac{x-1}{x+1}\right) - 1} \]Simplify each part of the fraction separately. For the numerator:\[ \left(\frac{x-1}{x+1}\right) + 1 = \frac{x-1}{x+1} + \frac{x+1}{x+1} = \frac{(x-1) + (x+1)}{x+1} = \frac{2x}{x+1} \]For the denominator:\[ \left(\frac{x-1}{x+1}\right) - 1 = \frac{x-1}{x+1} - \frac{x+1}{x+1} = \frac{(x-1) - (x+1)}{x+1} = \frac{-2}{x+1} \]

Step 4: Simplify the Final Expression

Now substitute the simplified numerator and denominator back into the formula:\[ G\left(\frac{x-1}{x+1}\right) = \frac{\frac{2x}{x+1}}{\frac{-2}{x+1}} \]Since the fractions have a common denominator, simplify it:\[ G\left(\frac{x-1}{x+1}\right) = \frac{2x}{-2} = -x \]

Step 5: Determine the Value at Specific Input, if Needed

If a specific input \(x\) is given or implied for the final result, substitute it into the simplified expression \(-x\). Otherwise, the general form is \(-x\).

Key Concepts

Reciprocal Functions

Reciprocal functions are a crucial concept in mathematics, especially when dealing with rational functions. The term "reciprocal" in general refers to the inverse of a number. For a function, specifically, taking the reciprocal means this: if you have a function represented as \( f(x) = \frac{a}{b} \), then its reciprocal function is \( f^{-1}(x) = \frac{b}{a} \). This is significant in various mathematical contexts, particularly in solving equations and composing functions.

In the provided solution, we worked with the function \( G(x) = \frac{x+1}{x-1} \). Its reciprocal function is then \( \frac{1}{G(x)} \) or \( \frac{x-1}{x+1} \). Knowing how to find the reciprocal is essential to carry out the function composition effectively, as seen in the problem statement. Always keep in mind that finding reciprocity in functions helps not only in function composition but also provides insights into function behavior regarding zeroes and undefined points.

Simplifying Expressions

Simplifying expressions is a fundamental skill in algebra. It involves altering the expression into its simplest form without changing its value. This process can help make equations easier to solve or interpret, especially in more complex operations like function compositions.

In our function composition \( G\left(\frac{1}{G(x)}\right) \), we used simplification heavily. When you replace \( x \) with \( \frac{x-1}{x+1} \) in \( G(x) \), obtaining a simplified form of the expression can make the process efficient and less prone to errors.

When simplifying:

• Combine like terms or factors whenever possible.
• Use arithmetic operations carefully to avoid mistakes, especially with negatives and fractions.
• For fractions, identify common denominators or factors to combine terms.

Through simplification, the final expression \( G\left(\frac{x-1}{x+1}\right) \) became \(-x\). This step highlights how simplifying expressions is not just about numerical ease but also conceptual clarity.

Piecewise Functions

Piecewise functions are defined by different expressions based on varying intervals of the input value \( x \). While they weren't directly employed in this exercise, understanding them can deepen comprehension of function behavior and help apply similar logic in managing complex functions.

Consider a piecewise function which behaves differently over different intervals of \( x \), such as:

• If \( x < 0 \), then \( f(x) = 2x + 1 \)
• If \( x \geq 0 \), then \( f(x) = x^2 \)

When working with composed functions or reciprocals, think about how a piecewise function might behave if \( G^{-1}(x) \) shifts as \( x \) moves through these intervals. This can illustrate concepts such as:

• Differing outputs from a single input based on defined conditions.
• Importance of domain and range in ensuring function definability across intended intervals.

Even if not used explicitly, the conceptual understanding of piecewise functions offers a framework that can assist in effectively managing and visualizing pieces of functions in complex compositions.