Problem 55 - A Graphical Approach to Precalculus with Limits

Question

Use \(f(x)\) and \(g(x)\) to find each composition. Identify is domain. (Use a calculator if necessary to find the domain.) \(\begin{array}{llll}\text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x)\end{array}\) $$f(x)=\frac{1}{x+1}, g(x)=5 x$$

Step-by-Step Solution

Answer

(a) \((f \circ g)(x)=\frac{1}{5x+1}\), domain: all reals except \(-\frac{1}{5}\). (b) \((g \circ f)(x)=\frac{5}{x+1}\), domain: all reals except \(-1\). (c) \((f \circ f)(x)=\frac{x+1}{x+2}\), domain: all reals except \(-1, -2\).

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

The expression \((f \circ g)(x)\) means that we take the output of \(g(x)\) and use it as the input for \(f(x)\).Start by substituting \(g(x) = 5x\) into \(f(x) = \frac{1}{x+1}\). This gives us: \[(f \circ g)(x) = f(g(x)) = f(5x) = \frac{1}{5x+1}\].The domain of \(f(x)\) requires that the denominator is not zero. So, we need to solve:\[5x + 1 eq 0\].Subtract 1:\[5x eq -1\].Divide by 5:\[x eq -\frac{1}{5}\].Thus, the domain of \((f \circ g)(x)\) is all real numbers except \(x = -\frac{1}{5}\).

Step 2: Determine \((g \circ f)(x)\)

The expression \((g \circ f)(x)\) means take the output of \(f(x)\) and use it as the input for \(g(x)\).Substitute \(f(x) = \frac{1}{x+1}\) into \(g(x) = 5x\). This gives us: \[(g \circ f)(x) = g(f(x)) = g\left(\frac{1}{x+1}\right) = 5\left(\frac{1}{x+1}\right) = \frac{5}{x+1}\].The domain of \((g \circ f)(x)\) requires that the denominator is not zero, so solve:\[x + 1 eq 0\].Subtract 1:\[x eq -1\].Hence, the domain of \((g \circ f)(x)\) is all real numbers except \(x = -1\).

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

The expression \((f \circ f)(x)\) implies inputting \(f(x)\) back into itself.Substitute \(f(x) = \frac{1}{x+1}\) into itself: \[(f \circ f)(x) = f(f(x)) = f\left(\frac{1}{x+1}\right) = \frac{1}{\left(\frac{1}{x+1}\right)+1}\].Simplify by finding a common denominator inside the expression:\[= \frac{1}{\left(\frac{1}{x+1} + \frac{x+1}{x+1}\right)} = \frac{1}{\frac{x+2}{x+1}} = \frac{x+1}{x+2}\].To find the domain, we need both the initial \(f(x)\) and the simplified expression to have non-zero denominators. This gives two conditions:1. \(x+1 eq 0\), so \(x eq -1\).2. \(x+2 eq 0\), so \(x eq -2\).Therefore, the domain of \((f \circ f)(x)\) is all real numbers except \(x = -1\) and \(x = -2\).

Key Concepts

Domain of Functions

Understanding the domain of a function is a crucial concept in mathematics. It refers to all the possible input values (usually 'x' values) for which a function is defined. For instance, if we're dealing with a function that involves division, the denominator cannot be zero as division by zero is undefined. To find the domain, we look at the expression and identify any values that could cause issues like negative square roots or zeros in a denominator.

When dealing with a function like \(f(x) = \frac{1}{x+1}\), the domain is all real numbers except when the denominator equals zero. By setting \x+1 eq 0\, we find that \x eq -1\. Thus, \(f(x)\) is defined for all real numbers except \x = -1\. This principle is applied consistently when checking the domain for composite functions as well.

Rational Functions

Rational functions are expressions that are the ratio of two polynomials, much like fractions. In a rational function \(f(x) = \frac{p(x)}{q(x)}\), \p(x)\ and \q(x)\ are polynomials, and the domain is all real numbers except where \q(x) = 0\.

For example, with \(f(x) = \frac{1}{x+1}\), this is a simple rational function where the denominator cannot be zero. Solving \x + 1 eq 0\ gives \x eq -1\, imposing a restriction on the domain.

Rational functions can become more interesting when they involve composition like \((f \circ g)(x)\). Here, you'd have another rational form like \(\frac{1}{5x+1}\), requiring you to re-evaluate the domain by ensuring the new denominator \5x+1\ is also not zero.

Composite Functions

Composite functions combine two functions where the output of one becomes the input of the other. This is noted as \(f \circ g\)(x) = f(g(x))\. For effective evaluation and finding domains, substitute the inner function into the outer one.

To compose \((f \circ g)(x)\), take \(g(x)\) and insert it into \(f(x)\). For example, with \(f(x) = \frac{1}{x+1}\) and \(g(x) = 5x\), the composition becomes \(\frac{1}{5x+1}\). This requires ensuring the new expression doesn't result in zero denominators.

Similarly, explore \((g \circ f)(x)\), which would be \(\frac{5}{x+1}\), necessitating \x + 1 eq 0\. Recognizing these compositions and their unique domains is crucial for deeper understanding and application in mathematics.