Problem 73 - Calculus Single Variable

Question

Let \(f(x)=x^{p}\) for some fixed power \(p .\) Is it ever true that \(f \circ f=f \cdot f ?\)

Step-by-Step Solution

Answer

Yes, when \( p = 0 \) or \( p = 2 \), \( f \circ f = f \cdot f \).

Step 1: Understanding the Composition of Functions

The composition of the function \( f(x) = x^p \) with itself, \( (f \circ f)(x) \), means that we apply \( f \) to the result of \( f(x) \). Therefore, \( (f \circ f)(x) = f(f(x)) = f(x^p) = (x^p)^p = x^{p^2} \).

Step 2: Understanding the Multiplication of Functions

The multiplication of the function \( f(x) = x^p \) with itself, \( f \cdot f \), means that we multiply \( f(x) \) by \( f(x) \). Thus, \( (f \cdot f)(x) = f(x) \cdot f(x) = x^p \cdot x^p = x^{2p} \).

Step 3: Equating the Composition and Multiplication

To find if \( f \circ f = f \cdot f \), we set \( x^{p^2} = x^{2p} \). For these powers to be equal for all \( x \), their exponents must be equal: \( p^2 = 2p \).

Step 4: Solving the Equation for p

Rearrange the equation \( p^2 = 2p \) to \( p^2 - 2p = 0 \). Factor out \( p \), giving \( p(p - 2) = 0 \). This means \( p = 0 \) or \( p = 2 \).

Step 5: Checking the Solutions

If \( p = 0 \), then \( f(x) = x^0 = 1 \) for all \( x eq 0 \). Both sides \( 1 \cdot 1 = 1 \) and \( 1 = 1 \) are trivially equal. If \( p = 2 \), then \( f(x) = x^2 \), and both the composition \( (x^2)^2 = x^4 \) and multiplication \( x^2 \cdot x^2 = x^4 \) are equal, satisfying the condition.

Key Concepts

Function Composition

Function composition is a concept where two functions are combined to form a new function. Consider the function \( f(x) = x^p \). When we compose this function with itself, written as \( f \circ f \), we plug the entire function \( f(x) \) into itself again.
Here's a breakdown:

• The notation \( (f \circ f)(x) \) is the same as \( f(f(x)) \).
• If \( f(x) = x^p \), then \( f(x^p) = (x^p)^p \).
• This results in \( (f \circ f)(x) = x^{p^2} \).

Breaking down a function this way can help us understand how inputs transform as they pass through sequential functions. This is a cornerstone of calculus, offering insights into how functions relate when connected together.

Function Multiplication

Function multiplication concerns multiplying two functions to produce a new one. If we multiply the function \( f(x) = x^p \) by itself, we express this as \( f \cdot f \). Let's see how this works in practice:

• In simpler terms, \( (f \cdot f)(x) \) means \( f(x) \times f(x) \).
• Plugging in the function: \( x^p \times x^p \).
• According to the laws of exponents, \( x^p \times x^p = x^{2p} \).

Function multiplication expands on standard algebraic operations by applying them in the context of functions. This arithmetic combines output values, creating meaningful new expressions you will often encounter in calculus and algebra.

Power Functions

Power functions are an essential part of algebra and calculus, characterized by an expression in the form \( f(x) = x^p \). Here, \( p \) is a constant that determines the function's behavior.
Let's identify some key features:

• Positive Powers: When \( p \) is positive, the function increases as \( x \) increases, resulting in a curve that rises or falls depending on the value of \( p \).
• Zero Power: If \( p = 0 \), the function simplifies to \( f(x) = 1 \), since any number raised to zero is 1.
• Importance: Power functions serve as the foundation for more complicated functions, forming the basis of polynomial expressions.

These functions are pivotal in modeling natural phenomena where relationships can be expressed as powers of variables, making them useful in many real-world applications.

Algebraic Equations

Algebraic equations involve expressions set equal to each other and are solved to find unknown values. In our context, we solve for the power \( p \).
Consider the given equation \( p^2 = 2p \). To find the values of \( p \) that satisfy the equality, we can manipulate the equation:

• First, rewrite it as \( p^2 - 2p = 0 \).
• Factor the equation: \( p(p - 2) = 0 \).
• The solutions are when either \( p = 0 \) or \( p - 2 = 0 \), so \( p = 2 \).

Solving algebraic equations often involves techniques such as factoring, substitution, or using the quadratic formula. Solving equations like these is fundamental for calculus and many areas of mathematics, helping to uncover values that make an expression true.