Problem 75 - Calculus Single Variable

Question

Let \(f(x)=x^{5}, I(x)=x,\) and \(p(x)=\pi\) for all \(x .\) Find functions \(\varphi, \psi, \mu\) and \(\lambda\) such that \(\varphi \circ f=p, f \circ \psi=p, \mu \circ f=I+p\) and \(f \circ \lambda=I+p\)

Step-by-Step Solution

Answer

\(\varphi(x) = \pi\), \(\psi(x) = \pi^{1/5}\), \(\mu(x) = x^{1/5} + \pi\), \(\lambda(x) = (x + \pi)^{1/5}\).

Step 1: Understand the Given Functions

We have three functions defined as follows: \(f(x) = x^5\), \(I(x) = x\), and a constant function \(p(x) = \pi\). Our task is to find functions \(\varphi, \psi, \mu,\) and \(\lambda\) that satisfy specific function compositions with \(f, I,\) and \(p.\)

Step 2: Solve for \(\varphi\)

We need to determine \(\varphi\) such that \(\varphi \circ f = p\). This can be written as \(\varphi(f(x)) = \pi\). Since \(f(x) = x^5\), this implies \(\varphi(x^5) = \pi\). Thus, \(\varphi\) must be a function that outputs \(\pi\) for any input, therefore \(\varphi(x) = \pi\) for all \(x.\)

Step 3: Solve for \(\psi\)

We need \(f \circ \psi = p\), or equivalently \(f(\psi(x)) = \pi\). Since \(f(y) = y^5\), we want \(\psi(x)^5 = \pi\). Therefore, \(\psi(x) = \pi^{1/5}\) for each \(x.\)

Step 4: Solve for \(\mu\)

Find \(\mu\) such that \(\mu \circ f = I + p\). This means \(\mu(f(x)) = x + \pi\). Since \(f(x) = x^5\), this becomes \(\mu(x^5) = x + \pi\). Hence, \(\mu(y) = y^{1/5} + \pi\) for all \(y.\)

Step 5: Solve for \(\lambda\)

We need to find \(\lambda\) such that \(f \circ \lambda = I + p\), i.e., \(f(\lambda(x)) = x + \pi\). That implies \(\lambda(x)^5 = x + \pi\). Thus, \(\lambda(x) = (x + \pi)^{1/5}\) for all \(x.\)

Key Concepts

Inverse Functions

Inverse functions essentially help us reverse operations done by a function. If we have a function \( f(x) = y \), then the inverse function \( f^{-1}(y) \) will return the original value \( x \). In other words, a function and its inverse undo each other.

• **Existence of Inverse:** To have an inverse, a function must be one-to-one (bijective). This means it produces a unique output for every input.
• **Finding Inverses:** Generally, we swap \( x \) and \( y \) in \( y = f(x) \) to \( x = f(y) \) and solve for \( y \).

In the context of the exercise, by reversing the effect of \( f(x) = x^5 \) with different compositions, we find specific inverse-like functions:- **For \( \psi(x) \):** \( f(\psi(x)) = \pi \) implies \( \psi(x) = \pi^{1/5} \). This essentially reverses the power of 5 for a constant output.
- **For \( \lambda(x) \):** \( f \circ \lambda = I + p \) gives \( \lambda(x) = (x + \pi)^{1/5} \), relating the operation to a specific input incremented by \( \pi \).

Constant Functions

Constant functions produce a fixed output no matter the input. A typical example is a function \( g(x) = c \), where \( c \) is a constant number. No matter which value is substituted for \( x \), the result will always remain \( c \).

In solving function composition problems, constant functions are pivotal because they simplify complexity:

• A constant function's graph is always a horizontal line as its output does not vary.
• In compositions like \( \varphi \circ f = p \), the constant output \( \varphi(x) = \pi \) helps meet the requirement because its simplicity means \( \varphi \) ignores the input.

This trait ensures simplicity when dealing with function compositions in complex equations. Constant functions like \( p(x) = \pi \) are especially useful when paired with other functions for ensuring a stable condition is met across all scenarios.

Exponential Functions

Exponential functions assume the form \( f(x) = a^x \), where \( a \) is a positive constant greater than 1. These functions grow extremely quickly because they increase by a constant percentage over equal increments.

• **Characteristic Behavior:** Unlike linear growth, exponential functions increase more rapidly. For instance, the growth from 2 to 3 is dramatic as we move forward on the function's curve.
• **Inverse Relation:** The exponential function's inverse is typically a logarithmic function, assisting in dialing back the rapid changes induced by exponentiation.

In the context of the exercise involving \( f(x) = x^5 \), while this is technically a polynomial, it showcases exponential-like growth characteristics seen in power functions. Manipulating this with inverses such as \( \psi(x) = \pi^{1/5} \) effectively undoes the rapid growth, demonstrating potency similar to that of actual exponential functions. This demonstrates how manipulation of powers (exponents) can intricately affect the behavior of functions and their compositions.