Problem 73 - 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 \(g \circ f\). $$f(x)=\frac{1}{x} ; g(x)=2 x+5$$

Step-by-Step Solution

Answer

\((f \circ g)(x) = f(g(x)) = \frac{1}{(2x+5)},\) with a domain of all real numbers different from \(x=-\frac{5}{2}\). And \((g \circ f)(x) = g(f(x)) = \frac{2}{x} + 5,\) where the domain is all real numbers except \(x=0\).

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

To do this, substitute \(g(x)\) into \(f(x)\) in place of x. Thus, \((f \circ g)(x) = f(g(x)) = f(2x+5) = \frac{1}{(2x+5)}.\)

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

The function \(\frac{1}{(2x+5)}\) is defined for every real number except for \(x=-\frac{5}{2}\), since a division by zero is undefined. Therefore, the domain is all real numbers different from \(x=-\frac{5}{2}\).

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

This time, substitute \(f(x)\) into \(g(x)\) in x's place. Thus, \((g \circ f)(x) = g(f(x)) = g(\frac{1}{x}) = 2(\frac{1}{x})+5 = \frac{2}{x} + 5.\)

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

For the function \(\frac{2}{x} + 5\), it's undefined when \(x=0\), because a division by zero is undefined. Therefore, the domain is all real numbers except \(x=0\).

Key Concepts

Function Composition

Understanding function composition is a fundamental concept in algebra that involves combining two functions into a single function. When we have two different functions, say f(x) and g(x), composing them involves applying one function to the result of another.

For instance, the composition (f \(\circ\) g)(x) means you first evaluate g(x), and then you apply the function f to that result. If we look at the given exercise, to find (f \(\circ\) g)(x), we substitute g(x), which is 2x+5, into f, leading to f(2x+5), and hence to our composed function \(\frac{1}{2x+5}\). It's like a function within a function, or a mathematical 'nesting doll'.

The reverse composition, (g \(\circ\) f)(x), is the other way around. We start with f(x), which is \(\frac{1}{x}\), and plug it into g. So g\left(\frac{1}{x}\right) results in \frac{2}{x} + 5. It's essential to follow the correct order since function composition is not necessarily commutative; f \(\circ\) g can be different from g \(\circ\) f.

Domain of a Function

The domain of a function refers to all the possible input values (or 'x' values) that will output real numbers when substituted into a function. It is essentially the set of 'allowed' inputs.

In the context of our exercise, determining the domain for the composite functions (f \(\circ\) g) and (g \(\circ\) f) requires us to consider where each inner function is defined and where the resulting composition makes sense. For \(\frac{1}{2x+5}\), our function is undefined at x=-\frac{5}{2} because that would result in a zero denominator, which is not permitted in mathematics. Thus, the domain of (f \(\circ\) g) is all real numbers except x=-\frac{5}{2}.

Similarly, with the function \(\frac{2}{x} + 5\), the problematic input is x=0 because it also leads to division by zero. Therefore, the domain of (g \(\circ\) f) excludes zero, encompassing all real numbers except for x=0. Recognizing the domain ensures that a function's evaluation is always mathematically valid.

Inverse Operations

Inverse operations are mathematical actions that undo each other, such as addition and subtraction or multiplication and division. When we look at functions, finding an inverse function means finding a function that 'reverses' the effect of the original function.

For a function f(x), its inverse f^{-1}(x) will return the original input when applied to the output of f. That is, if y=f(x), then x=f^{-1}(y). It's indispensable for solving equations since it allows us to isolate the variable we're solving for. The exercise we are examining doesn't specifically deal with finding inverse functions, but the interplay of operations in function composition shares conceptual similarities with inverses.

While inverse functions are about 'undoing', and function composition is about 'doing' in sequence, both concepts are critical ways of linking functions together to describe complex relationships and require a deep understanding of how functions behave under various operations.