Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

For \(\Phi(u)=u^{3}+1\) and \(\Psi(v)=1 / v\), find each value. ( \(\Psi\) is the uppercase Greek letter psi.) (a) \((\Phi+\Psi)(t)\) (b) \((\Phi \circ \Psi)(r)\) (c) \((\Psi \circ \Phi)(r)\) (d) \(\Phi^{3}(z)\) (e) \((\Phi-\Psi)(5 t)\) (f) \(((\Phi-\Psi) \circ \Psi)(t)\)

Step-by-Step Solution

Answer

(a) \(t^3 + 1 + \frac{1}{t}\), (b) \(\frac{1}{r^3} + 1\), (c) \(\frac{1}{r^3 + 1}\), (d) \((z^3 + 1)^3\), (e) \(125t^3 + 1 - \frac{1}{5t}\), (f) \(\frac{1}{t^3} + 1 - t\)."

Step 1: Understanding the Functions

We are given two functions: \( \Phi(u) = u^3 + 1 \) and \( \Psi(v) = \frac{1}{v} \). Our task is to find the values for different combinations and compositions of these functions.

Step 2: Solve (Φ+Ψ)(t)

Calculate \((\Phi+\Psi)(t)\) by adding the functions. \( \Phi(t) = t^3 + 1 \) and \( \Psi(t) = \frac{1}{t} \). Thus, \( (\Phi + \Psi)(t) = t^3 + 1 + \frac{1}{t} \).

Step 3: Solve (Φ∘Ψ)(r)

Find \((\Phi \circ \Psi)(r)\), which means substitute \(\Psi(r)\) into \(\Phi(u)\):\(\Psi(r) = \frac{1}{r} \)So, \( \Phi(\Psi(r)) = \Phi\left(\frac{1}{r}\right) = \left(\frac{1}{r}\right)^3 + 1 = \frac{1}{r^3} + 1 \).

Step 4: Solve (Ψ∘Φ)(r)

Find \((\Psi \circ \Phi)(r)\), substitute \(\Phi(r)\) into \(\Psi(v)\):\( \Phi(r) = r^3 + 1 \)So, \( \Psi(\Phi(r)) = \Psi(r^3 + 1) = \frac{1}{r^3 + 1} \).

Step 5: Solve Φ³(z)

Find \(\Phi^3(z)\), which is \(\Phi(z)\) raised to the third power:\(\Phi(z) = z^3 + 1\)So \(\Phi^3(z) = (z^3 + 1)^3\).

Step 6: Solve (Φ-Ψ)(5t)

Find \((\Phi-\Psi)(5t)\) by subtracting the functions: \( \Phi(5t) = (5t)^3 + 1 = 125t^3 + 1 \) and \( \Psi(5t) = \frac{1}{5t} \). Thus, \( (\Phi - \Psi)(5t) = 125t^3 + 1 - \frac{1}{5t} \).

Step 7: Solve ((Φ-Ψ)∘Ψ)(t)

Find \(((\Phi-\Psi)\circ \Psi)(t)\) by substituting \(\Psi(t)\) into \((\Phi-\Psi)\): \( \Psi(t) = \frac{1}{t} \) \( \Phi\left(\frac{1}{t}\right) = \left(\frac{1}{t}\right)^3 + 1 = \frac{1}{t^3} + 1 \) \( \Psi\left(\frac{1}{t}\right) = t \) So, \((\Phi-\Psi)(\Psi(t)) = \left(\frac{1}{t^3} + 1 - t\right)\).

Key Concepts

Function Addition

Function addition is the process of summing two functions to create a new one. If you have two functions, say \( f(x) \) and \( g(x) \), their sum \( (f + g)(x) \) is simply \( f(x) + g(x) \). Applying this to our example, consider \( \Phi(t) = t^3 + 1 \) and \( \Psi(t) = \frac{1}{t} \). The addition of these two functions will be \( (\Phi + \Psi)(t) = t^3 + 1 + \frac{1}{t} \).

This process of function addition is vital in many areas of mathematics and engineering, where combining different function outputs into a single expression can simplify complex operations or models.

• Helps in integrating multiple function behaviors into one unified output
• Simplifies complex equations by reducing them to a single expression
• Used in fields like physics, economics for modeling purposes

Function Subtraction

Function subtraction involves subtracting one function from another. If we have functions \( f(x) \) and \( g(x) \), then their difference, denoted by \( (f - g)(x) \), is \( f(x) - g(x) \). From the original exercise, we have \( \Phi(5t) = 125t^3 + 1 \) and \( \Psi(5t) = \frac{1}{5t} \), so \((\Phi - \Psi)(5t) = 125t^3 + 1 - \frac{1}{5t} \).

Subtraction of functions is as important as their addition and is frequently used to determine the change or difference between two datasets or theories.

• Aids in comparing different models or sets of data
• Helps in analyzing the rate of change in physical quantities
• Crucial for computations in calculus and differential equations

Power Function

A power function is one where potential expressions involve exponents. The general form is \( f(x) = x^n \), where \( n \) is any real number. In the given exercise, \( \Phi(z) = z^3 + 1 \) is considered a power function because its primary variable \( z \) is raised to the power of 3.

Moreover, when we calculate \( \Phi^3(z) \), we raise the entire expression of \( \Phi(z) \) to the third power: \((z^3 + 1)^3\).

Power functions appear in polynomial equations, are used for growth models, and help describe behaviors depending on natural laws like gravity or elasticity.

• Essential for understanding polynomial expressions
• Used widely in natural sciences and economics
• Describes scalable patterns in datasets

Inverse Function

An inverse function "undoes" the operation of a particular function. If you have a function \( f(x) \) and its inverse \( f^{-1}(x) \), then \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). In our exercise, \( \Psi(v) = \frac{1}{v} \) is essentially finding the multiplicative inverse of \( v \).

While the given functions \( \Phi \) and \( \Psi \) are not each other's inverses, the composition \( (\Psi \circ \Phi)(r) = \frac{1}{r^3 + 1} \) manipulates the input through inverse-related operations.

Inverse functions are foundational in trigonometry, calculus, and algebra, often used to solve equations where the desired variable is embedded within a function.

• Important for reversing operations or solving equations
• Utilized to simplify complex mathematical equations
• Helps in understanding and deriving functions from datasets