Problem 74 - 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 \(g \circ f\). $$f(x)=3 x+1 ; g(x)=\frac{2}{x}$$

Step-by-Step Solution

Answer

The expressions for \(f \circ g\) and \(g \circ f\) are \(6/x + 1\) and \(2/ (3x + 1)\), respectively. The domain of \(f \circ g\) is all real numbers excluding 0 and the domain of \(g \circ f\) is all real numbers excluding -1/3.

Step 1: Find the Expressions for \(f \circ g\) and \(g \circ f\)

To find \(f \circ g\), we substitute \(g(x)\) into \(f(x)\) to find \(f(g(x))\). Therefore, \(f(g(x)) = 3 (2/x) + 1 = 6/x + 1\). Similarly, to find \(g \circ f\), we substitute \(f(x)\) into \(g(x)\) to find \(g(f(x))\). Therefore, \(g(f(x)) = 2/ (3x + 1)\).

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

The domain of \(f \circ g\) is the set of all x-values for which \(6/x + 1\) is defined. This occurs for all x-values except 0. Hence, the domain of \(f \circ g\) is all real numbers excluding 0. The domain of \(g \circ f\) is the set of all x-values for which \(2/ (3x + 1)\) is defined. This occurs for all x-values except -1/3. Hence, the domain of \(g \circ f\) is all real numbers excluding -1/3.

Key Concepts

Function Composition

Understanding function composition is crucial for grasping more complex mathematical ideas in precalculus. Let's simplify it. Essentially, composing two functions is like blending their processes. Think of it as putting the output of one function straight into the next. For example, when we combine the functions f(x) and g(x), denoted as (f \(circ\) g)(x), we're doing just that.

Domains of Composite Functions

The domain of a composite function can be a tricky beast, but don't worry, we've got this! It tells us which inputs are 'legal' for our function operation. For our composite function (f \(circ\) g)(x), we've got to make sure that g(x) is valid (since it happens first), and then that f(g(x)) keeps things on the straight and narrow. It's like a two-step verification process: if g(x) gives us an 'illegal' output, f won't take it. This is why the composite function (f \(circ\) g) excludes x=0, since feeding a zero into g(x) causes trouble (you can't divide by zero!).

Inverse Functions

Imagine if functions had twins - but the kind that does everything in reverse. Enter inverse functions. They're the mathematical equivalent of 'Control + Z'. When a function takes an input and gives an output, its inverse takes that output and returns to the original input. Symbolically, if we have a function f, its inverse is denoted as f^{-1}. Now, not all functions are lucky enough to have an inverse. To qualify, a function must be one-to-one — each output comes from one unique input — otherwise, the 'undo' process wouldn't know where to go back to.