Problem 84 - 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 \(g \circ f\). $$f(x)=|x| ; g(x)=\frac{x^{2}+3}{x^{2}-4}$$

Step-by-Step Solution

Answer

\(f \circ g(x) = \left| \frac{x^{2}+3}{x^{2}-4} \right|\) and its domain is all real numbers except -2 and 2. \(g \circ f(x) = \frac{x^2+3}{x^2-4}\) and its domain is all real numbers except -2 and 2.

Step 1: Finding Expression and Domain for \(f \circ g(x)\)

First, substitute \(g(x)\) into \(f(x)\) for \(f \circ g(x)\): \n\n \(f \circ g(x) =f(g(x))=|g(x)|= \left|\frac{x^{2}+3}{x^{2}-4}\right|\) \n\n The domain of \(f \circ g(x)\) is all real values of x except for those that make the denominator of the rational function \(g(x)\) equal to zero - because division by zero is undefined. So, let's find those values: \n \n \(x^2 - 4 =0\ => x^2 = 4 \ => x=2, -2\) \n\n Therefore, The domain of \(f \circ g\) is all real numbers except \(-2, 2\)

Step 2: Finding Expression and Domain for \(g \circ f(x)\)

Next, substitute \(f(x)\) into \(g(x)\) for \(g \circ f(x)\): \n \n \(g \circ f(x) =g(f(x))=\frac{f(x)^{2}+3}{f(x)^{2}-4}=\frac{|x|^2+3}{|x|^2-4}=\frac{x^2+3}{x^2-4}\) \n\nNotice that for \(g\circ f(x)\) we got the same function as \(g(x)\) since squaring the absolute value of \(x\) is equal to squaring \(x\). \n\nTherefore, the domain of \(g \circ f(x)\) is also all real numbers except \(-2, 2\)

Key Concepts

Function Composition

Understanding function composition is crucial in mathematics as it combines two or more functions to form a new function. When composing two functions, say \(f \) and \(g\), the output of \(g\) becomes the input for \(f\). This is represented by \(f \circ g\), read as 'f composed with g'.

For instance, if we have \(f(x) = |x|\) and \(g(x) = \frac{x^2 + 3}{x^2 - 4}\), composing them into \(f \circ g(x)\) involves substituting \(g(x)\) into \(f\), which results in the expression \(\left|\frac{x^2 + 3}{x^2 - 4}\right|\). It's vital to consider the domain of the new function, as not all inputs from \(g\) might be valid inputs for \(f\).

Domain of a Function

The domain of a function is the collection of all permissible input values (typically represented by \(x\)) for which the function is defined. Identifying the domain involves determining the set of real numbers that don't cause any undefined operations—like division by zero—within the function.

Looking at \(g(x) = \frac{x^2 + 3}{x^2 - 4}\), we find the domain by setting its denominator equal to zero and solving for \(x\), yielding the values that are excluded from the domain. Since \(x^2 - 4 = 0\) leads to \(x = 2\) and \(x = -2\), we exclude these from the domain, resulting in all real numbers except 2 and -2.

Absolute Value Function

An absolute value function, often notated as \(f(x) = |x|\), takes a real number \(x\) and outputs its non-negative value. The function is visualized as a 'V' shape in a graph, with its vertex at the origin (0,0).

It's used within composition to ensure non-negativity, as in \(f \circ g(x) = |g(x)|\). In the given exercise, \(f\) is an absolute value function. When used in function composition, such as in \(f \circ g(x)\), it 'absorbs' the sign of \(g(x)\), effectively removing any negative values from the result.

Rational Function

A rational function is a ratio of two polynomials, written in the form \(g(x) = \frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials and the denominator, \(Q(x)\), is not equal to zero. Since division by zero is undefined, the values that make \(Q(x)\) zero are excluded from the domain.

In our exercise, \(g(x)\) is a rational function, \(\frac{x^2 + 3}{x^2 - 4}\). To fully grasp the function's behavior, we not only determine its domain but also recognize how it is altered when composed with other functions. For instance, \(g \circ f(x)\) again forms a rational function that maintains the same domain restrictions due to the squared absolute value of \(x\), thus excluding the same inputs of 2 and -2.