Problem 12 - Calculus with Concepts in Calculus

Question

Let \(f(x)=\frac{x-1}{x^{2}+1}\) and \(g(x)=x^{1 / 4}\). Find the specified values. $$ (f / g)\left(\frac{1}{4}\right) $$

Step-by-Step Solution

Answer

\((f / g)\left(\frac{1}{4}\right) = \frac{-12 \sqrt{2}}{17}\).

Step 1: Understand the Composite Function

We need to find \( (f / g)\left(\frac{1}{4}\right) \), which means \( \frac{f\left(\frac{1}{4}\right)}{g\left(\frac{1}{4}\right)} \). This involves calculating both \( f\left(\frac{1}{4}\right) \) and \( g\left(\frac{1}{4}\right) \) first, then dividing \( f\left(\frac{1}{4}\right) \) by \( g\left(\frac{1}{4}\right) \).

Step 2: Calculate \( f\left(\frac{1}{4}\right) \)

Substitute \( x = \frac{1}{4} \) into \( f(x) = \frac{x-1}{x^{2}+1} \):\[f\left(\frac{1}{4}\right) = \frac{\frac{1}{4} - 1}{\left(\frac{1}{4}\right)^{2} + 1}\]This simplifies to:\[f\left(\frac{1}{4}\right) = \frac{-\frac{3}{4}}{\frac{1}{16} + 1} = \frac{-\frac{3}{4}}{\frac{1}{16} + \frac{16}{16}} = \frac{-\frac{3}{4}}{\frac{17}{16}} = \frac{-3 \times 16}{4 \times 17} = \frac{-48}{68} = \frac{-12}{17}\]

Step 3: Calculate \( g\left(\frac{1}{4}\right) \)

Substitute \( x = \frac{1}{4} \) into \( g(x) = x^{1/4} \):\[g\left(\frac{1}{4}\right) = \left(\frac{1}{4}\right)^{1/4} = \frac{1}{\sqrt{2}} \approx 0.7071\]

Step 4: Compute \( \frac{f\left(\frac{1}{4}\right)}{g\left(\frac{1}{4}\right)} \)

Now divide the values obtained from \( f\left(\frac{1}{4}\right) \) and \( g\left(\frac{1}{4}\right) \):\[(f / g)\left(\frac{1}{4}\right) = \frac{-12/17}{1/\sqrt{2}} = \frac{-12 \times \sqrt{2}}{17}\]

Key Concepts

Composite Functions

Composite functions involve the combination of two different functions. When you're asked to find \( (f/g)(x) \), the operation is not exactly the same as with composite functions like \( f(g(x)) \), but it can be conceptually similar, as it involves managing the interactions between two functions.
In this exercise, you're required to evaluate the function \( f(x) \) and the function \( g(x) \) separately before executing the division process. The concept primarily prepares you to understand how two functions influence the value of a third function together.
Composite functions help in:

• Breaking down complex expressions into simpler parts.
• Providing a clearer overview of function behavior over an interval.
• Facilitating advanced calculus operations, such as integration and differentiation.

The steps for dealing with problems like these start with first evaluating the individual functions with the given input.This offers an essential foundation in understanding how inputs pass through multiple layers of functions in different scenarios.

Function Division

Function division involves dividing one function by another, and it's a critical operation in calculus.When we talk about \( (f/g)(x) \), we're specifically referencing the division of \( f(x) \) by \( g(x) \),which is essentially performing an operation similar to finding the quotient of two numbers. However, there is an added layer of complexity because given functions may have varying behaviors at different parts of their domains.
Here are some tips to keep in mind with function division:

• Always ensure \( g(x) \) is not zero, as division by zero is undefined.
• Watch out for restrictions in the domains of both functions.
• Simplify numerators and denominators where possible, as it might ease the calculation.

In our example, calculating \( (f/g)(1/4) \) meant first plugging \( 1/4 \) into both \( f(x) \) and \( g(x)\), then simplifying each individual result before final division.This ensures clarity and accuracy when looking at complex expressions like those involving polynomial and radical components.

Radicals in Functions

Functions that involve radicals can add a layer of complexity due to the nature of roots.Radicals require careful handling, especially when they appear in the denominator or when raised to fractional powers.In this exercise, the function \( g(x) = x^{1/4} \) demonstrates the use of radicals by converting a fraction to a root.
It's key to remember:

• Roots imply both a principal root and sometimes a need to consider negative values.However, depending on the context (e.g., square roots in \( \mathbb{R} \)), only non-negative roots might be taken as non-imaginary.
• Handling radicals often requires simplification techniques, like rationalizing the denominator, to make calculations manageable.
• When executing arithmetic operations with radicals, it's crucial to keep track of square roots, cube roots, etc.,so you don't miscalculate by inadvertently changing the essence of the original term.

By breaking down the expression \( g(1/4) = (1/4)^{1/4} \), we observed that the function generally less than its input value, reflecting the fact that roots reduce the magnitude of positive numbers.