Problem 84 - College Algebra

Question

For the following exercises, use \(f(x)=x^{3}+1\) and \(g(x)=\sqrt[3]{x-1}\). Let \(f(x)=\frac{1}{x}\). a. Find \((f \circ f)(x)\). b. Is \((f \circ f)(x)\) for any function \(f\) the same result as the answer to part (a) for any function? Explain.

Step-by-Step Solution

Answer

(f ◦ f)(x) = x. This result is unique to specific functions like f(x) = 1/x.

Step 1: Simplify the Function f(x)

We are given that \( f(x) = \frac{1}{x} \). This is a rational function where the variable \( x \) is in the denominator, making the function undefined at \( x = 0 \).

Step 2: Understand the Composition of Functions (f ◦ f)(x)

The composition \((f \circ f)(x)\) means plugging \( f(x) \) into itself. This can be expressed as \( f(f(x)) \).

Step 3: Compute (f ◦ f)(x)

First, compute \( f(x) \) which is \( \frac{1}{x} \). Now substitute \( \frac{1}{x} \) back into \( f(x) \), yielding \( f(f(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} \). This simplifies to \( x \), since dividing by a fraction is equivalent to multiplying by its reciprocal.

Step 4: Simplification Result

Thus, we find that \( (f \circ f)(x) = x \), demonstrating that composing the function \( f(x) = \frac{1}{x} \) with itself results in the identity function.

Step 5: Generalization of (f ◦ f) for Any Function

The result \((f \circ f)(x) = x \) is unique to functions that act as their own inverses, like \( f(x) = \frac{1}{x} \). In general, \( (f \circ f)(x) \) is not necessarily equal to \( x \) for all functions. The resulting function depends on the nature of \( f(x) \).

Key Concepts

Rational Functions

Rational functions are a type of function where the output is the result of a quotient involving polynomial expressions. For example, the function given in the exercise: \( f(x) = \frac{1}{x} \) is a rational function. Here, the variable \( x \) is in the denominator of the fraction. This placement is crucial as it influences where the function is defined.

The value at \( x = 0 \) is of particular importance because it causes the function to be undefined. Division by zero does not produce a valid result in mathematics, thus \( f(x) \) is undefined at this point. When considering rational functions, always check for undefined points, which will relate to the domain of the function.

• Identifying undefined spots is essential for determining the domain.
• Polynomials in both the numerator and the denominator influence the function shape.

In summary, rational functions combine polynomial expressions in a ratio, with their domains being limited by points where the denominator equals zero.

Identity Function

Identity functions are special kinds of functions where the output is exactly equal to the input value. The classic form of an identity function is \( f(x) = x \). Applying the identity function to any input simply gives back the same input.

In the exercise, when \( (f \circ f)(x) = x \), we demonstrate that plugging the function into itself results in this identity behavior. For the function \( f(x) = \frac{1}{x} \), this means \( f(f(x)) = x \). Therefore, composing the function with itself effectively neutralizes any transformation stacked on the initial input.

• The role of identity functions is often to simplify or check if operations cancel each other out.
• It shows how some functions, through composition, can revert back to the beginning value.

The observation here that \( f \circ f \) results in the identity function is a valuable insight in function operations, illustrating simplicity among seemingly complex inputs.

Function Inverses

Function inverses are concepts where two functions essentially "undo" each other. If you possess a function \( f \) and its inverse \( g \), then the compositions \( f(g(x)) \) and \( g(f(x)) \) both yield the identity function, \( x \).

In the exercise, the function \( f(x) = \frac{1}{x} \) acts as its own inverse because when it is composed with itself, it results in the identity function \( f(f(x)) = x \). This behavior is typical of simple rational functions and shows that function inverses can occasionally be the same function.

• Inverse functions reveal important symmetrical properties.
• A function is its own inverse if its composition equals the identity function.

Understanding function inverses is fundamental to grasping more complex operations, as it assures mutual cancellation in the composite process.

Domain of Functions

The domain of a function is the set of input values for which the function is defined. For rational functions such as \( f(x) = \frac{1}{x} \), care must be taken because they have restrictions on the input values due to division by zero.

In the exercise, the domain of \( f(x) \) excludes \( x = 0 \), as it would cause the function to be undefined. When composing functions like \( (f \circ f)(x) \), it's essential to determine if any further restrictions appear on the domain due to the operations within the composition.

• When excluding values in the domain, consider any variable in the denominator.
• Composition can alter or maintain domain restrictions of the original functions.

Always articulate and confirm domains when dealing with complex compositions to ensure the results are valid across expected input values.