Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(f(s)=\sqrt{s^{2}-4}\) and \(g(w)=|1+w|\), find formulas for \((f \circ g)(x)\) and \((g \circ f)(x)\).

Step-by-Step Solution

Answer

\((f \circ g)(x) = \sqrt{(1 + x)^{2} - 4}\) and \((g \circ f)(x) = |1 + \sqrt{x^{2} - 4}|\)."

Step 1: Understanding Composition of Functions

The composition of functions involves applying one function to the results of another. For two functions, say \( f(x) \) and \( g(x) \), the composition \( (f \circ g)(x) \) is defined as \( f(g(x)) \). Similarly, \( (g \circ f)(x) \) is defined as \( g(f(x)) \). We need to apply these definitions to \(f(s)=\sqrt{s^{2}-4}\) and \(g(w)=|1+w|\).

Step 2: Finding \((f \,\circ\, g)(x)\)

To find \((f \,\circ\, g)(x)\), substitute \(g(x)\) into \(f(s)\). Start with \(g(x) = |1 + x|\).Now compute \(f(g(x)) = f(|1 + x|)\).Substitute \(|1+x|\) into \(f(s)\):\[ f(|1+x|) = \sqrt{(|1+x|)^{2} - 4} \].Since \(|1+x|^{2} = (1+x)^{2}\), we have:\[ f(|1+x|) = \sqrt{(1 + x)^{2} - 4} \].

Step 3: Finding \((g \,\circ\, f)(x)\)

To find \((g \,\circ\, f)(x)\), substitute \(f(x)\) into \(g(w)\).Start with \(f(x) = \sqrt{x^{2} - 4}\).Now compute \(g(f(x)) = g(\sqrt{x^{2} - 4})\).Substitute \(\sqrt{x^2 - 4}\) into \(g(w)\):\[ g(\sqrt{x^{2} - 4}) = |1 + \sqrt{x^{2} - 4}| \].

Step 4: Final Composition Formulas

After substituting and simplifying, we find the two compositions of the functions: - \((f \,\circ\, g)(x) = \sqrt{(1 + x)^{2} - 4}\).- \((g \,\circ\, f)(x) = |1 + \sqrt{x^{2} - 4}|\).

Key Concepts

Understanding Function Composition

Function composition is a process that involves two or more functions being combined to form a new function. Imagine you have two functions: an outer function, denoted as \( f(x) \), and an inner function, denoted as \( g(x) \). The composition of these functions is expressed as \((f \circ g)(x)\), which reads as "\( f \) of \( g(x) \)." This notation means you first apply \( g(x) \) and then use its result as the input for \( f(x) \).
In simple terms,

• Evaluate the inner function: Start with \( g(x) \).
• Use the result as input: Feed the output of \( g(x) \) into \( f(x) \).
• Conclude the result: The final answer is \( f(g(x)) \).

This operation might seem complex at first but is similar to combining ingredients in a recipe to create a dish. First, you prepare the separate components and then blend them to come up with a new flavor profile.

Exploring Square Roots

The concept of square roots is essential in mathematics, especially when dealing with functions like \( f(s) = \sqrt{s^2 - 4} \). A square root asks the question, "what number, when multiplied by itself, gives you the original number?". For example, the square root of 9 is 3, because \(3 \times 3 = 9\).
When you see a term like \( \sqrt{s^2 - 4} \), part of what you’re calculating is ensuring that the expression inside the root—under the radical—is non-negative. This means \( s^2 - 4 \geq 0 \). Understanding how to handle square roots and the requirements for their expressions helps avoid negative results under the square-root sign, which cannot be simplified in the real number system.

Demystifying Absolute Value

Absolute value is a simple yet powerful concept represented by two vertical bars surrounding a number or expression, like \(|1 + w|\). The absolute value of a number gives its distance from zero on the number line, always as a non-negative value. It essentially ignores any negative sign in front of a number.
For instance,

• \(|5| = 5\)
• \(|-3| = 3\)

Absolute values are crucial in various calculations, such as when dealing with

• Distance: Always a non-negative quantity.
• Expressions: Making complex situations easier.
• In composition: Simplifying nested functions.

In the function \(g(w) = |1 + w|\), no matter whether \(w\) is positive or negative, you always evaluate \(1 + w\) and then take its absolute value.

Applying the Substitution Method

The substitution method is pivotal when dealing with function composition. It involves replacing parts of one function with another function, allowing you to solve for or simplify complicated expressions. This technique is valuable in managing compositions like \((f \circ g)(x)\) and \((g \circ f)(x)\).
Let's break down how substitution works:

• Identify what to replace: Decide which part of a function (like \( g(x) \) in \( f(g(x)) \)) needs substitution.
• Make the substitution: Replace the identified part with the corresponding expression. For example, replace \( g(x) \) in \( f(x) \) with its formula \( |1 + x| \).
• Simplify: Calculate step-by-step to reach the final composition results.

This method simplifies complex compositions and makes them easier to manage by breaking them down into understandable parts.