Problem 40 - Precalculus with Limits

Question

In Exercises 37-40, find (a) \(f \circ g\), (b) \(g \circ f\), and (c) \(g \circ g\). \(f(x) = x^3\), \(g(x) = \frac{1}{x}\)

Step-by-Step Solution

Answer

The answers are as follows: (a) \(f \circ g= \frac{1}{x^3}\), (b) \(g \circ f= \frac{1}{x^3}\), and (c) \(g \circ g= x\)

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

To find \(f \circ g\), we substitute \(g(x)\) into \(f(x)\). Given \(f(x) = x^3\) and \(g(x) = \frac{1}{x}\), we replace \(x\) in the equation for \(f\) with \(g(x)\) to get \(f(g(x)) = (\frac{1}{x})^3 = \frac{1}{x^3}\).

Step 2: Compute \(g \circ f\)

To compute \(g \circ f\), we substitute \(f(x)\) into \(g(x)\). With \(f(x) = x^3\) and \(g(x) = \frac{1}{x}\), we substitute \(x\) in the equation for \(g\) with \(f(x)\) resulting in \(g(f(x)) = \frac{1}{{(x^3)}}\).

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

Last of all, to compute \(g \circ g\) we substitute \(g(x)\) into \(g(x)\) itself. Using the function \(g(x) = \frac{1}{x}\), when we replace \(x\) in the equation for \(g\) with \(g(x)\), we get \(g(g(x)) = \frac{1}{{\frac{1}{x}}} = x\).

Key Concepts

Composite Functions: Combining Two Functions Together

Composite functions are all about putting one function inside another. Think of it like layers in a sandwich. When you work with composite functions, you take the output from one function and use it as an input for another. In mathematics, this is usually represented by the symbol \( \circ \). Here's how you can understand composite functions better:

• When you see \( f \circ g \), it means you need to take the function \( g(x) \) and put it inside \( f(x) \). In our example, this results in \( f(g(x)) = (\frac{1}{x})^3 = \frac{1}{x^3} \).
• Conversely, \( g \circ f \) implies taking the function \( f(x) \) and placing it inside \( g(x) \). This results in \( g(f(x)) = \frac{1}{x^3} \).

The order of functions is crucial. \( f \circ g \) is not the same as \( g \circ f \), highlighting how order affects the final result. Think of it as a different sequence of actions in a process, each leading to a unique outcome.

Inverse Functions: Undoing What a Function Does

Inverse functions are like the reverse magic tricks of mathematics. They take the result of a function and work backwards to find the original input. If \( f(x) \) transforms input \( x \) into output \( y \), then the inverse function, denoted \( f^{-1}(x) \), transforms \( y \) back to \( x \). Important points about inverse functions include:

• For an inverse to exist, a function must be one-to-one, meaning each output derives from exactly one input.
• In our example with \( f(x) = x^3 \) and \( g(x) = \frac{1}{x} \), neither of these functions are inverses of each other, but they can be inverted separately.
• An example is the function \( g(x) = \frac{1}{x} \), which is its own inverse, because applying it twice results in the original \( x \), as seen in \( g(g(x)) = x \).

Knowing the inverse helps check work, determine an input, or solve equations, making them a powerful tool in algebra and beyond.

Function Operations: Working With Functions Like Numbers

Function operations involve doing arithmetic with functions, much like numbers. You can add, subtract, multiply, or divide functions. Ensuring operations are done correctly is crucial. Here's a breakdown:

• Addition \((f + g)(x)\): Combine outputs of \( f(x) \) and \( g(x) \) by adding them together.
• Subtraction \((f - g)(x)\): Calculate the difference between \( f(x) \) and \( g(x) \).
• Multiplication \((f \cdot g)(x)\): Multiply the outputs of \( f(x) \) and \( g(x) \). It expands how functions can interact.
• Division \((\frac{f}{g})(x)\): Divide \( f(x) \) by \( g(x) \), but watch for \( g(x) eq 0 \), which keeps things defined.

Performing these operations helps navigate through complex mathematical problems where multiple functions appear together. By understanding how to work with them, you can simplify expressions and solve varied equations efficiently.