Problem 75 - 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 \(g \circ f\). $$f(x)=\frac{3}{2 x+1} ; g(x)=2 x^{2}$$

Step-by-Step Solution

Answer

\(f(g(x)) = \frac{3}{4x^{2} + 1}\) with domain of \(x \in R\), and \(g(f(x)) = \frac{18}{(2x + 1)^{2}}\) with domain \(x > -1/2\)

Step 1 : Compute \(f \circ g\)

Substitute \(x\) in \(f(x)\) by \(g(x)\). That is, \(f(g(x)) = f\left(2x^{2}\right) = \frac{3}{2\left(2x^{2}\right) + 1} = \frac{3}{4x^{2} + 1}\

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

For the domain of \(f \circ g(x)\), it's the set of all 'x' in the domain of 'g', such that \(g(x)\) is in the domain of 'f'. The denominator cannot be zero, so solve the inequality \(4x^2 + 1 > 0\), which is true for all real numbers. Therefore, the domain of \(f \circ g(x)\) is \(x \in R\)

Step 3 : Compute \(g \circ f\)

Substitute \(x\) in \(g(x)\) by \(f(x)\). That is, \(g(f(x)) = g\left(\frac{3}{2x + 1}\right) = 2 \left(\frac{3}{2x + 1}\right)^2 = \frac{18}{\left(2x +1\right)^2}\

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

For the domain of \(g \circ f(x)\), it's the set of all 'x' in the domain of 'f', such that \(f(x)\) is in the domain of 'g'. The denominator cannot be zero, so solve the inequality \(2x + 1 > 0\), which results in \(x > -1/2\). Therefore, the domain of \(g \circ f(x)\) is \(x \in R\) and \(x > -1/2\)

Key Concepts

Domain of Functions

Understanding the domain of a function is crucial when working with mathematical expressions. The domain of a function is the set of all possible input values (usually represented by 'x') for which the function is defined. In simple terms, it refers to all the values that can be plugged into the function without causing any mathematical issues, such as dividing by zero or taking the square root of a negative number.

In our exercise, we have two functions:

• \(f(x) = \frac{3}{2x+1}\)
• \(g(x) = 2x^2\)

Let's focus on the domain of each. For \(f(x)\), the expression is a rational function, which means you need to ensure the denominator \(2x + 1\) is not zero. Thus, the domain excludes the value of \(x\) that makes \(2x + 1 = 0\), or \(x = -\frac{1}{2}\). For \(g(x)\), since it is a simple quadratic function \(2x^2\) without any potential division by zero or negative square roots, its domain is all real numbers \(x \in \mathbb{R}\).

In composite functions, understanding the domain requires ensuring that each composed part remains valid. This ensures the resulting expression is meaningful and error-free.

Composite Functions

Composite functions involve combining two functions into one new function. This is a process of substituting one function into another. If we have two functions, \(f(x)\) and \(g(x)\), the composition \((f \circ g)(x)\) is read as "f of g of x" and defined as \(f(g(x))\). It means we're plugging the entire function \(g(x)\) into \(f(x)\) in place of \(x\).

In the given exercise, we calculated:

• \((f \circ g)(x) = \frac{3}{4x^2 + 1}\), where the expression from \(g(x)\) (\(2x^2\)) replaces 'x' in the function \(f\).
• \((g \circ f)(x) = \frac{18}{(2x + 1)^2}\), where the expression from \(f(x)\) (\(\frac{3}{2x + 1}\)) is input into function \(g\).

The tricky part of finding the composition is ensuring each step results in a valid function expression. This means the output of \(g(x)\) should be a valid input for \(f(x)\), and vice versa for \((g \circ f)(x)\). It's important to closely observe any restrictions in the original functions' domains that might impact the composition.

Rational Functions

Rational functions are composed of ratios of polynomials. They often involve expressions where one polynomial is divided by another, like \(\frac{a}{b}\), with both 'a' and 'b' as polynomial expressions. These functions are important in mathematics because they model many real-world situations and appear frequently in calculus.

The key aspect to consider with rational functions is their domain. Specifically, you need to identify any values of the variable that would make the denominator equal zero, as these would make the function undefined. Thus, rational functions have specific restrictions in their domain. For the function \(f(x) = \frac{3}{2x+1}\), as seen in the exercise, \(2x + 1\) should not be zero. Solving \(2x + 1 = 0\) gives \(x = -\frac{1}{2}\), a point where the function is undefined.

Rational functions can present additional challenges when they form part of a composite function, such as when we found \((f \circ g)(x)\) and \((g \circ f)(x)\). Extra care is required to ensure restrictions from these compositions are respected and accurately determine their domains.