Problem 45 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=\frac{1}{x}+x\) and \(g(x)=\frac{2 x}{x^{2}+1} .\) Evaluate the following expressions. (a) \(g\left(f\left(\frac{1}{3}\right)\right)\) (b) \(f\left(g\left(\frac{1}{3}\right)\right)\)

Step-by-Step Solution

Answer

The evaluations of the expressions are given by (a) \( g(f (\frac{1}{3}))=\frac{12}{37} \) and (b) \( f(g (\frac{1}{3}))=\frac{34}{15} \)

Step 1: Calculate \( f(\frac{1}{3}) \)

Substitute \( x = \frac{1}{3} \) into the function \( f(x) = \frac{1}{x} + x \). That gives \( f(\frac{1}{3})= \frac{1}{\frac{1}{3}}+ \frac{1}{3}=3+3=6 \)

Step 2: Calculate \( g(\frac{1}{3}) \)

Substitute \( x = \frac{1}{3} \) into the function \( g(x) = \frac{2x}{x^2 + 1} \). That gives \( g(\frac{1}{3})= \frac{2*1/3}{(1/3)^2 + 1}= \frac{2/3}{1/9 + 1}= \frac{2/3}{10/9}=\frac{3}{5} \)

Step 3: Calculate \( g(f(\frac{1}{3})) \)

Now, substitute \( x = 6 \) (the result from step 1) into function \( g(x) \). This gives us \( g(6)= \frac{2*6}{6^2 + 1}=\frac{12}{36+1}= \frac{12}{37} \). So, \( g(f(\frac{1}{3}))=\frac{12}{37} \)

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

Finally, substitute \( x = \frac{3}{5} \) (the result from step 2) into function \( f(x) \). This gives us \( f(\frac{3}{5})= \frac{1}{3/5}+\frac{3}{5}= \frac{5}{3}+\frac{3}{5}= \frac{25+9}{15}= \frac{34}{15} \). So, \( f(g(\frac{1}{3}))=\frac{34}{15} \)

Key Concepts

Rational Functions

Rational functions are mathematical expressions that represent the ratio of two polynomials. They come in the general form of \(R(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials. The function is defined as long as \(Q(x) eq 0\), because division by zero is undefined. Rational functions can exhibit a wide range of behaviors, such as intercepts, asymptotes, and holes. These characteristics occur where the function is either undefined or the numerator becomes zero.

In our exercise, both \(f(x) = \frac{1}{x} + x\) and \(g(x) = \frac{2x}{x^2 + 1}\) are examples of rational functions. The function \(f(x)\) is interesting because it involves a simple rational part, \(\frac{1}{x}\), added to a linear function, \(x\). Meanwhile, \(g(x)\) is a rational function that includes a quadratic expression in the denominator \(x^2 + 1\). Understanding how each component contributes to the overall behavior of the function is crucial for analyzing these functions fully.

Function Evaluation

Function evaluation is the process of calculating the output of a function for a specific input. To evaluate a function, you simply substitute the given value of the variable into the function's expression and simplify.

In the provided exercise, we evaluated \(f\) and \(g\) for \(x = \frac{1}{3}\). For \(f(x)\), this involved substituting \(\frac{1}{3}\) into \(\frac{1}{x} + x\), resulting in 6. Similarly, for \(g(x)\), we substitute \(\frac{1}{3}\) into \(\frac{2x}{x^2 + 1}\), giving us \(\frac{3}{5}\).

These evaluations were then used to determine \(g(f(\frac{1}{3}))\) and \(f(g(\frac{1}{3}))\). Each of these compositions required a subsequent evaluation of one function with the result of the other. Proper function evaluation is crucial for accurately finding these compositions.

Mathematical Expressions

Mathematical expressions are combinations of numbers, variables, operators, and functions constructed according to the rules of mathematics. They don't include an equality sign or inequality symbol which differentiates them from equations. Understanding expressions involves recognizing the components and how they are related or operate together.

In this context, the expressions for \(f(x)\) and \(g(x)\) need to be manipulated correctly during substitution and simplification. For example:

• When evaluating \(f(\frac{1}{3})\), convert \(\frac{1}{\frac{1}{3}}\) to a whole number first, which simplifies to 3, then add \(\frac{1}{3}\).
• Similarly, for \(g(\frac{1}{3})\), simplifying fractional expressions like \(\frac{2/3}{10/9}\) relies on multiplying by the reciprocal.

Breaking down and understanding mathematical expressions allows for correct evaluation and manipulation, essential skills in solving composition of function problems like those presented in this exercise.