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) \(\left(\begin{array}{lll}\Phi & \circ & \Psi\end{array}\right)(r)\) (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) \(\frac{1}{r^3 + 1}\); (e) \(125t^3 + 1 - \frac{1}{5t}\); (f) \(\frac{1}{t^3} + 1 - t\). \(\Phi^3(z)\) is complex to compute directly.

Step 1: Understanding Functions

First, identify the given functions: \( \Phi(u) = u^3 + 1 \) and \( \Psi(v) = \frac{1}{v} \). We will use these to perform various operations.

Step 2: Part (a): Sum of Functions \\(\Phi+\Psi\\)

To find \((\Phi+\Psi)(t)\), compute it as \( \Phi(t) + \Psi(t) \). Substituting the expressions, we have:\[\Phi(t) = t^3 + 1, \quad \Psi(t) = \frac{1}{t}\]Thus, \((\Phi + \Psi)(t) = t^3 + 1 + \frac{1}{t}\).

Step 3: Part (b): Composition \\(\Phi \circ \Psi\\)

To find \((\Phi \circ \Psi)(r)\), substitute \( \Psi(r) \) into \( \Phi(u) \). Since \( \Psi(r) = \frac{1}{r} \), we have:\[\Phi(\Psi(r)) = \Phi\left(\frac{1}{r}\right) = \left(\frac{1}{r}\right)^3 + 1 = \frac{1}{r^3} + 1\]

Step 4: Part (c): Composition \\(\Psi \circ \Phi\\)

To find \((\Psi \circ \Phi)(r)\), substitute \( \Phi(r) \) into \( \Psi(v) \). With \( \Phi(r) = r^3 + 1 \), the expression is:\[\Psi(\Phi(r)) = \Psi(r^3 + 1) = \frac{1}{r^3 + 1}\]

Step 5: Part (d): Power of Function \\(\Phi^3\\)

To compute \(\Phi^3(z)\), you take the function \(\Phi\) composed with itself three times: \(\Phi(\Phi(\Phi(z)))\). Calculate as follows:\[\Phi(z) = z^3 + 1\]Calculate \( \Phi(\Phi(z)) \) and then \( \Phi(\Phi(\Phi(z))) \); this is complex and typically you'll just be expected to set it up due to the rapidly expanding algebra.

Step 6: Part (e): Difference of Functions \\(\Phi-\Psi\\)

For \((\Phi - \Psi)(5t)\), compute:\[\Phi(5t) = (5t)^3 + 1 = 125t^3 + 1, \quad \Psi(5t) = \frac{1}{5t}\]Thus, \((\Phi - \Psi)(5t) = 125t^3 + 1 - \frac{1}{5t}\).

Step 7: Part (f): Nested Composition \\(\Phi-\Psi \) and \\ \Psi\\

To find \(((\Phi - \Psi) \circ \Psi)(t)\), calculate \(\Phi(\Psi(t))\) and \(\Psi(\Psi(t))\) first, then subtract:1. \( \Psi(t) = \frac{1}{t}\)2. \(\Phi\left(\frac{1}{t}\right) = \left(\frac{1}{t}\right)^3 + 1 = \frac{1}{t^3} + 1\)3. \(\Psi\left(\frac{1}{t}\right) = t\)Thus, \(((\Phi - \Psi) \circ \Psi)(t) = \left(\frac{1}{t^3} + 1\right) - t = \frac{1}{t^3} + 1 - t\).

Key Concepts

Function Operations

Function operations refer to ways of combining two or more functions to create a new function. The most common operations include addition, subtraction, multiplication, and division. For instance, if you have two functions, say \(f(x)\) and \(g(x)\), you can create:

• Sum: \((f + g)(x) = f(x) + g(x)\)
• Difference: \((f - g)(x) = f(x) - g(x)\)
• Product: \((f \cdot g)(x) = f(x) \cdot g(x)\)
• Quotient: \((f / g)(x) = \frac{f(x)}{g(x)}\) if \(g(x) eq 0\)

Understanding how these basic operations work can help solve more complex problems. For example, if you want to compute \((\Phi + \Psi)(t)\), you use the mentioned addition rule: plug in \(t\) in both \(\Phi(t)\) and \(\Psi(t)\), and then add the results together. Similarly, for \((\Phi - \Psi)(5t)\), evaluate each function at \(5t\) before subtracting the results.

Function Composition

Function composition involves the application of one function to the results of another function, effectively creating a "function of a function." It is denoted as \((f \circ g)(x)\), which reads as \(f\) of \(g(x)\). Here, \(g\) is applied first, and then \(f\) is applied to the result of \(g\). For example:

• \((\Phi \circ \Psi)(r)\): First find \(\Psi(r)\), which is \(\frac{1}{r}\), then plug this into \(\Phi\), resulting in \(\Phi\left(\frac{1}{r}\right)\) which becomes \(\frac{1}{r^3} + 1\).
• \((\Psi \circ \Phi)(r)\): Find \(\Phi(r) = r^3 + 1\) first, and then substitute this into \(\Psi\), giving \(\frac{1}{r^3+1}\).

Function composition helps to break complex problems into simpler parts, essentially tackling one function at a time.

Function Powers

Function powers are an interesting concept where a function is applied to its result multiple times, denoted as \(f^n(x)\), where \(n\) indicates the number of times the function is applied. For instance, \(f^2(x) = f(f(x))\) and so on. It's crucial to note that this is different from raising a function's output to a power. In the exercise, \(\Phi^3(z)\) means applying \(\Phi\) three times in a row:

• Calculate \(\Phi(z)\) as \(z^3 + 1\).
• Apply \(\Phi\) again: \(\Phi(\Phi(z)) = \Phi(z^3 + 1)\).
• The process continues, involving lengthy calculations, highlighting the rapid increase in complexity with each application.

Approaching function powers methodically is essential, ensuring each step is clearly worked out to avoid mistakes in computation.