Problem 69 - Intermediate Algebra

Question

Let \(f(x)=\frac{1}{x}\) and \(g(x)=\frac{1}{x^{2}} .\) Find each of the following. $$ (g \circ f)\left(\frac{1}{3}\right) $$

Step-by-Step Solution

Answer

\((g \circ f)\left(\frac{1}{3}\right) = \frac{1}{9}\).

Step 1: Understand the Composition

The symbol \((g \circ f)(x)\) means that you first apply the function \(f\) to \(x\), and then apply the function \(g\) to the result. Thus, \((g \circ f)(x) = g(f(x))\).

Step 2: Find \(f\left(\frac{1}{3}\right)\)

Substitute \(\frac{1}{3}\) into the function \(f(x) = \frac{1}{x}\):\[ f\left(\frac{1}{3}\right) = \frac{1}{\frac{1}{3}} = 3 \]

Step 3: Apply \(g(x)\) to the Result of \(f(x)\)

Now that we know \(f\left(\frac{1}{3}\right) = 3\), substitute this into \(g(x) = \frac{1}{x^2}\):\[ g(f(x)) = g(3) = \frac{1}{3^2} = \frac{1}{9} \]

Step 4: Write Down the Final Result

The value of \((g \circ f)\left(\frac{1}{3}\right)\) is the result from applying \(g\) to \(f\left(\frac{1}{3}\right)\), which we calculated to be \(\frac{1}{9}\).

Key Concepts

Inverse Functions

Inverse functions are like the reverse route on a map – they take you back to where you started. If a function maps an input to an output, its inverse goes the other way, mapping that output back to the input. Not all functions have inverses, but when they do, we denote the inverse of a function \( f \) as \( f^{-1} \).

To find an inverse function, you essentially swap the roles of inputs and outputs. For example, if \( f(x) = y \), then in the inverse function, \( f^{-1}(y) = x \). To identify if a function has an inverse, you need to check if it's one-to-one, which means that each output is the result of exactly one input.

Remember, a fascinating property of inverse functions is that they "undo" each other, such as when \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). This property is invaluable in mathematics and helps solve a wide range of problems by reversing processes.

Function Evaluation

Function evaluation is like putting information into a vending machine to get a snack in return. You input a value into a function, and the function provides you with a corresponding result. Mathematically, function evaluation involves substituting the input value into the function's formula.

For instance, if we have a function \( f(x) = \frac{1}{x} \) and want to evaluate it at \( x = \frac{1}{3} \), we replace \( x \) with \( \frac{1}{3} \) to get \( f\left(\frac{1}{3}\right) = \frac{1}{\frac{1}{3}} = 3 \). It’s like asking the function a specific question: "What do I get if I give you \( \frac{1}{3} \)?"

Function evaluation is crucial because it helps determine how functions behave under specific circumstances, enabling predictions and deeper understanding of mathematical models.

Composite Functions

Composite functions are like passing a message through a chain of people – each person whispers the message to the next. In mathematics, you have two or more functions connected such that the output of one function becomes the input for another. The notation \((g \circ f)(x)\) illustrates that \( f(x)\) is applied first, and its result is subsequently fed into \( g \).

For example, with \( f(x) = \frac{1}{x} \) and \( g(x) = \frac{1}{x^2} \), the composite function \((g \circ f)(x))\) means first compute \( f(x)\) and then use that result in \( g(x) \). In our exercise, \( f\left(\frac{1}{3}\right) = 3 \) and so \( g(3) = \frac{1}{3^2} = \frac{1}{9} \).

Composite functions are a powerful tool in math, allowing complex operations to be broken down into simpler steps. They help in designing systems and solving multifaceted problems by simplifying them into serial actions.