Problem 47 - Calculus Volume 1

Question

For the following exercises, for each pair of functions, find a. \((f \circ g)(x)\) and b. \((g \circ f)(x)\) Simplify the results. Find the domain of each of the results. $$ f(x)=\frac{3}{2 x+1}, g(x)=\frac{2}{x} $$

Step-by-Step Solution

Answer

\((f \circ g)(x) = \frac{3x}{4+x}\), domain: \(x \neq 0, -4\). \((g \circ f)(x) = \frac{4x+2}{3}\), domain: \(x \neq -\frac{1}{2}\).

Step 1: Find \((f \circ g)(x)\)

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). This gives us:\[(f \circ g)(x) = f(g(x)) = f\left(\frac{2}{x}\right) = \frac{3}{2 \left( \frac{2}{x} \right) + 1}\]Simplify the expression:\[(f \circ g)(x) = \frac{3}{\frac{4}{x} + 1} = \frac{3}{\frac{4+x}{x}} = \frac{3x}{4+x}\]

Step 2: Find the Domain of \((f \circ g)(x)\)

The domain of \((f \circ g)(x)\) is determined by the values of \(x\) such that both \(g(x)\) and the resulting function \(f(g(x))\) are defined.1. \(g(x)\) is undefined when \(x=0\) because \(g(x) = \frac{2}{x}\).2. \((f \circ g)(x)\) is also undefined when \(4+x = 0\) (denominator cannot be zero), i.e., \(x = -4\).Thus, the domain of \((f \circ g)(x)\) is \(x \in \mathbb{R} \setminus \{0, -4\}\).

Step 3: Find \((g \circ f)(x)\)

To find \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). This gives us:\[(g \circ f)(x) = g(f(x)) = g\left(\frac{3}{2x+1}\right) = \frac{2}{\frac{3}{2x+1}}\]Simplify the expression:\[(g \circ f)(x) = \frac{2(2x+1)}{3} = \frac{4x+2}{3}\]

Step 4: Find the Domain of \((g \circ f)(x)\)

The domain of \((g \circ f)(x)\) is determined by the values of \(x\) such that both \(f(x)\) and \(g(f(x))\) are defined.1. \(f(x)\) is undefined when \(2x+1=0\), i.e., \(x = -\frac{1}{2}\).Thus, the domain of \((g \circ f)(x)\) is \(x \in \mathbb{R} \setminus \{-\frac{1}{2}\}\).

Key Concepts

Domain of a Function

The domain of a function is the complete set of possible values of the independent variable, typically denoted as \(x\). This set includes all the values for which the function is defined. In more intuitive terms, the domain tells us the range of input values we can use for which our function will produce an output.

When working with function compositions, determining the domain involves examining each component function separately and then ensuring that the resulting composition remains defined over the overlapping valid inputs.

• In our given problem, for \(f(x) = \frac{3}{2x+1}\), the function is undefined when \(2x+1 = 0\) because division by zero is undefined. Solving for \(x\) gives us \(x = -\frac{1}{2}\).
• For \(g(x) = \frac{2}{x}\), it is undefined when \(x = 0\).

For the compositions, we need to check the domains once the functions are combined, making sure no new undefined points are introduced by the compositions themselves. For \(f \circ g(x)\), it is undefined both at \(x = 0\) and \(x = -4\) due to division restrictions.

Similarly, for \(g \circ f(x)\), the composition only introduces further restrictions by \(x = -\frac{1}{2}\). By considering these restrictions, we can correctly determine the valid inputs for each composite function.

Rational Functions

Rational functions are formed when one polynomial is divided by another. These functions typically resemble fractions, and their behavior is largely dictated by the denominator.

When working with rational functions, it's crucial to identify any points that would make the denominator zero, as these values create vertical asymptotes or undefined points within the function.

• Our function \(f(x) = \frac{3}{2x+1}\) is a perfect example of a rational function, where the numerator is the constant \(3\) and the denominator is the linear expression \(2x+1\).
• Similarly, \(g(x) = \frac{2}{x}\) is another rational function where the denominator is simply \(x\).

Their compositions, \(f \circ g(x) = \frac{3x}{4+x}\) and \(g \circ f(x) = \frac{4x+2}{3}\), remain rational functions because they apply the original rational function characteristics during their evaluation.

One of the key challenges in rational functions lies in determining these undefined values and simplifying the expressions to manage these characteristics effectively, ensuring they remain well-defined across their permissible domains.

Simplifying Expressions

Simplifying expressions involves reducing a complex mathematical expression into its simplest or most compact form. This is an essential skill that helps in understanding and solving equations more efficiently.

When we compose functions like \(f(g(x))\) or \(g(f(x))\), the initial expressions can often look complex. Simplifying them can provide clarity and reveal their true mathematical nature.

In our specific exercise, let's observe the simplification process:

• Initially, substituting \(g(x)\) into \(f(x)\) for \(f \circ g(x)\) gives us a more cumbersome expression like \((f \circ g)(x) = \frac{3}{\frac{4}{x} + 1}\).
• By multiplying through by \(x\) where appropriate, and combining terms, we simplify this to \(\frac{3x}{4+x}\).
• In the case of \(g \circ f(x)\), substituting yields \((g \circ f)(x) = \frac{2}{\frac{3}{2x+1}}\), which simplifies to a more manageable form, \(\frac{4x+2}{3}\).

Simplifying expressions not only makes them easier to work with but also helps in accurately determining domains and other key characteristics of the functions involved.