Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x)=\frac{1}{x-1}\). Find and simplify each value. (a) \(f(1 / x)\) (b) \(f(f(x))\) (c) \(f(1 / f(x))\)

Step-by-Step Solution

Answer

(a) \(\frac{x}{1-x}\), (b) \(-1\), (c) \(\frac{1}{x-2}\).

Step 1: Substitute x in f(x) with 1/x for f(1/x)

Given the function \(f(x) = \frac{1}{x-1}\), substitute \(x\) with \(\frac{1}{x}\). This gives: \[ f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x} - 1} \]This simplifies to: \[ f\left(\frac{1}{x}\right) = \frac{x}{1-x} \] by combining the fraction in the denominator.

Step 2: Substitute x in f(x) to find f(f(x))

Starting with \(f(x) = \frac{1}{x-1}\), substitute \(f(x)\) back into the function: \[ f(f(x)) = f\left(\frac{1}{x-1}\right) \]Using the result from Step 1, substitute \(\frac{1}{x-1}\) into the formula derived:\[ f(f(x)) = \frac{1}{\frac{1}{x-1} - 1} \]Simplify the denominator:\[ \frac{1}{x-1} - 1 = \frac{1-x}{x-1} \]This implies:\[ f(f(x)) = \frac{x-1}{1-x} = -1 \] after cancellation.

Step 3: Substitute 1/f(x) into f(x) to find f(1/f(x))

To find \(f\left(\frac{1}{f(x)}\right)\), calculate \(\frac{1}{f(x)}\): \[ \frac{1}{f(x)} = x-1 \]Then, substitute this back into \(f(x)\):\[ f(1/f(x)) = f(x-1) = \frac{1}{(x-1) - 1} = \frac{1}{x-2} \]

Key Concepts

Function Composition

When working with functions, sometimes we use the output of one function as the input for another. This is called function composition. Imagine function composition like a chain, where the output of one link becomes the input for the next.
For example, if we have two functions, such as our given \( f(x) = \frac{1}{x-1}\), composing two functions would look like \( f(g(x))\), where \( g(x) \) is a different function. In our case, we apply \( f(x) \) to itself, creating \( f(f(x)) \).
The steps involve substituting \( f(x) \) back into itself and simplifying it. This can help us understand how changing inputs affect the output when passed through multiple functions sequentially. It’s a useful skill for figuring out more complex systems in algebra.

Rational Functions

Rational functions are expressions made by dividing two polynomials. They are called 'rational' because they involve ratios. For instance, \( f(x) = \frac{1}{x-1} \) is a basic example. Here, the polynomial in the numerator is 1, and the denominator is \( x-1 \).
These functions often have interesting properties, like vertical asymptotes, which occur when the denominator equals zero. In our example, the function becomes undefined when \( x = 1 \), since division by zero isn't possible.
Rational functions can be quite quirky due to their undefined points, so it's important to understand how to approach them. Always watch out for values that make the denominator zero, and remember that simplifying these functions often involves factoring or finding common denominators.

Solving Rational Equations

To solve rational equations, the key is often simplification and finding a common denominator. These equations involve expressions where variables appear in the denominator, so handling them carefully is crucial.
Take our example function, \( f(x) = \frac{1}{x-1} \), and its transformed version \( f(\frac{1}{f(x)}) = \frac{1}{x-2} \). This involves substituting and inverse operations. Sometimes, like in part (c), you end up needing to work backwards by inverting fractions or clearing fractions to isolate variables.
Always check for any restricted values where the denominator could be zero. Practicing these steps enhances understanding and allows solving complex expressions with ease, avoiding potential pitfalls along the way.