Problem 85 - Precalculus : Building Concepts and Connections

Question

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=\frac{1}{x^{2}+1} ; g(x)=\frac{2 x+1}{3 x-1}$$

Step-by-Step Solution

Answer

The composite functions are \(f \circ g(x) = \frac{1}{\frac{13x^2-2x+2}{9x^2-2x+1}}\) with domain \(x ≠ \frac{1}{3}\), and \(g \circ f(x) = \frac{2y+1}{3y-1}\) with domain \(x ≠ ±\sqrt{2}\).

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

Firstly, replace each \(x\) in \(f(x)\) with \(g(x)\) to get: \(f \circ g(x) = f(g(x)) = \frac{1}{(g(x))^{2}+1} = \frac{1}{(\frac{2x+1}{3x-1})^{2}+1}\)

Step 2: Simplifying \(f \circ g(x)\) and finding its domain

Simplify the expression obtained in step 1 to its proper form. This involves squaring the fraction and adding 1 to the denominator. After simplifying, we obtain: \(f \circ g(x) = \frac{1}{(\frac{4x^2+4x+1}{9x^2-2x+1})+1} = \frac{1}{\frac{13x^2-2x+2}{9x^2-2x+1}}\). The domain for this function is all real numbers except where the denominator equals zero, hence the domain of \(f \circ g(x)\) is \(x ≠ \frac{1}{3}\)

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

Now we will find the composite function \(g \circ f(x)\) by replacing each \(x\) in \(g(x)\) with \(f(x)\): \(g \circ f(x) = g(f(x)) = \frac{2f(x)+1}{3f(x)-1} = \frac{2(\frac{1}{x^{2}+1})}{3(\frac{1}{x^{2}+1})-1}\)

Step 4: Simplifying \(g \circ f(x)\) and finding its domain

Simplify the expression obtained in step 3 to its proper form. This can be done by assuming \(y=x^{2}+1\) , simplifying, and substituting \(y\) back in. After simplifying, we obtain: \(g \circ f(x) = \frac{2y+1}{3y-1}\). The domain for this function is all real numbers except where the denominator equals zero. Hence, the domain of \(g \circ f(x)\) is \(x ≠ ±\sqrt{2}\)

Key Concepts

Domain of Functions

The domain of a function is essentially the set of all possible input values (often written as 'x-values') that can be safely used in a function without resulting in undefined or impossible numbers. For instance, you cannot divide by zero or take the square root of a negative number in the real number system.

To determine the domain of complex functions, you often need to consider restrictions from multiple functions involved. When a function is composed with another, like in composite functions, the domain will be further restricted. This means you must exclude any x-values that cause any part of the composition to be undefined.

• Look out for denominators. If a function has a fraction, the denominator cannot be zero.
• For square roots, all input under the root must be non-negative (if we are not considering complex numbers).

When dealing with rational or composite functions, it's a good practice to separately analyze each component before combining them.

Composite Functions

Composite functions occur when one function is applied to the result of another function. It's like a "function within a function". If you have two functions, say \( f \) and \( g \), a composite function is written as \( f \circ g \), which is read as "f composed with g".

To form a composite function, simply replace the input of the one function with another function. In the example given, \( f(x) \) is applied to \( g(x) \) to form \( f(g(x)) \) and vice versa for \( g(f(x)) \).

• Start by understanding each function individually before attempting to compose them.
• The resultant domain of the composite function must account for the restrictions from both original functions, as seen in the domains specified for \( f \circ g(x) \) and \( g \circ f(x) \).

The beauty of composite functions lies in seeing how one action impacts another, allowing us to solve more complex problems by breaking them into smaller parts.

Rational Functions

Rational functions are ratios of two polynomial functions, similar to how rational numbers are ratios of integers. For instance, you could say a rational function has the form \( R(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \).

A key feature of rational functions is understanding where they are undefined: anywhere \( Q(x) = 0 \). Calculating these points is essential for determining the domain and behavior of the function.

• Simplifying rational expressions often helps in identifying these critical x-values.
• In the step-by-step exercise, factorization and finding common denominators are useful techniques.

When working with rational functions, especially when involved in compositions, it's important to simplify and cautiously consider the domain. This ensures you understand where the function is valid and how it will behave when approaching restricted values.