Problem 47 - Calculus Volume 1

Question

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. \(\quad f(x)=\frac{3}{2 x+1}, g(x)=\frac{2}{x}\)

Step-by-Step Solution

Answer

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

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

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). Therefore, \((f \circ g)(x) = f(g(x)) = f\left(\frac{2}{x}\right)\). Substituting in, we get:\[ f\left(\frac{2}{x}\right) = \frac{3}{2\left(\frac{2}{x}\right) + 1} = \frac{3}{\frac{4}{x} + 1} \]To simplify, find a common denominator:\[ \frac{3}{\frac{4}{x} + 1} = \frac{3}{\frac{4}{x} + \frac{x}{x}} = \frac{3}{\frac{4 + x}{x}} = \frac{3x}{4 + x} \]

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

The expression \(\frac{3x}{4+x}\) implies that the domain excludes any \(x\) that makes the denominator zero. To find these values, set \(4+x = 0\):\[ x = -4 \]Additionally, since \(g(x)\) inputs into \(f(x)\), \(x\) must also not be zero (since \(g(x) = \frac{2}{x}\), and division by zero is undefined). Therefore, the domain of \((f \circ g)(x)\) is all real numbers except \(x = 0\) and \(x = -4\).

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

To find \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). Therefore, \((g \circ f)(x) = g(f(x)) = g\left(\frac{3}{2x+1}\right)\). Substituting in, we get:\[ g\left(\frac{3}{2x+1}\right) = \frac{2}{\left(\frac{3}{2x+1}\right)} = \frac{2(2x+1)}{3} = \frac{4x+2}{3} \]

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

The expression \(\frac{4x+2}{3}\) implies that \(f(x)\) must be defined. Thus, the denominator of \(f(x)\) cannot be zero, i.e., \(2x+1 eq 0\):\[ x eq -\frac{1}{2} \]Therefore, the domain of \((g \circ f)(x)\) is all real numbers except \(x = -\frac{1}{2}\).

Key Concepts

Composite Functions

Composite functions occur when you have two functions and combine them to create a third function. It involves substituting one function into another. When we say \((f \circ g)(x)\), we mean applying the function \(g(x)\) first, then using its result as the input for \(f(x)\). Similarly, \((g \circ f)(x)\) means applying \(f(x)\) first, then using it as the input for \(g(x)\).
To find a composite function, carefully substitute one function into the other.

• For \((f \circ g)(x)\), replace every instance of \(x\) in \(f(x)\) with \(g(x)\).
• For \((g \circ f)(x)\), do the opposite: replace every \(x\) in \(g(x)\) with \(f(x)\).

In the example provided, this means:

• \(f(x) = \frac{3}{2x+1}\) and \(g(x) = \frac{2}{x}\)
• \((f \circ g)(x) = \frac{3x}{4+x}\)
• \((g \circ f)(x) = \frac{4x+2}{3}\)

Simplifying these functions further helps in finding solutions and understanding the relationships between \(f\) and \(g\).

Domain of a Function

The domain of a function is the set of all possible inputs (or \(x\)-values) that the function can accept without giving an undefined result. When dealing with real-valued functions, the most common restriction in the domain occurs where the denominator is zero — since division by zero is undefined.
For composite functions like those in the example, you'd consider:

• Inputs that make any denominator zero.
• Any restrictions carried over from the inner function.

In \((f \circ g)(x)\), \(g(x)\) must be defined, which excludes \(x = 0\). And, \((f \circ g)(x)\) must not make its denominator zero, making \(x = -4\) another restriction.
Hence, the domain of \((f \circ g)(x)\) is all real numbers except \(x = 0\) and \(x = -4\).
For \((g \circ f)(x)\), we set the denominator of \(f(x)\) non-zero, preventing \(x = -\frac{1}{2}\). So, the domain is all reals except \(x = -\frac{1}{2}\).

Rational Functions

Rational functions are quotients of polynomial functions, resembling fractions, where the numerator and the denominator are polynomials. Commonly, rational functions appear in the form \(\frac{p(x)}{q(x)}\).
The awareness about rational functions is crucial because of their peculiar behavior at certain points — notably where \(q(x) = 0\).

• In \(\frac{3}{2x+1}\), \(2x+1 = 0\) implies a restriction at \(x = -\frac{1}{2}\).
• In \(\frac{2}{x}\), \(x = 0\) creates a restriction.

Rational functions often have interesting features:

• Vertical asymptotes, where the function value heads towards infinity as \(x\) approaches the critical value.
• Holes, a gap in the graph that's undefined at a specific point though it may appear otherwise continuous.

Understanding the nature of rational functions aids in predicting and explaining behaviors in composite functions or anywhere rationals are involved in calculations.