Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(f(x)=\sqrt{x^{2}-1}\) and \(g(x)=2 / x\), find formulas for the following and state their domains. (a) \((f \cdot g)(x)\) (b) \(f^{4}(x)+g^{4}(x)\) (c) \((f \circ g)(x)\) (d) \((g \circ f)(x)\)

Step-by-Step Solution

Answer

(a) \(\frac{2\sqrt{x^2 - 1}}{x}\), domain: \((-\infty, -1] \cup [1, \infty)\setminus\{0\}\). (b) \((x^2 - 1)^2 + \frac{16}{x^4}\), domain: \((-\infty, -1] \cup [1, \infty)\setminus\{0\}\). (c) \(\sqrt{\frac{4}{x^2} - 1}\), domain: \((-\infty, -2] \cup [2, \infty)\). (d) \(\frac{2}{\sqrt{x^2 - 1}}\), domain: \((-\infty, -1) \cup (1, \infty)\).

Step 1: Find the formula for \((f \cdot g)(x)\)

To find \((f \cdot g)(x)\), we multiply the functions \(f(x)\) and \(g(x)\). \[(f \cdot g)(x) = f(x) \cdot g(x) = \sqrt{x^2 - 1} \cdot \frac{2}{x} = \frac{2\sqrt{x^2 - 1}}{x}\].

Step 2: Determine the domain for \((f \cdot g)(x)\)

The domain is restricted by both individual functions. For \(f(x)\), \(x^2 - 1 \geq 0\) implies \(|x| \geq 1\). For \(g(x)\), \(x eq 0\). Thus, the domain of \((f \cdot g)(x)\) is \((-\infty, -1] \cup [1, \infty)\), excluding \(x=0\).

Step 3: Find the formula for \(f^{4}(x)+g^{4}(x)\)

First calculate \(f(x)^4\) and \(g(x)^4\): \[f^{4}(x) = (\sqrt{x^2 - 1})^4 = (x^2 - 1)^2\]\[g^{4}(x) = \left(\frac{2}{x}\right)^4 = \frac{16}{x^4}\] Thus, \[f^{4}(x) + g^{4}(x) = (x^2 - 1)^2 + \frac{16}{x^4}\].

Step 4: Determine the domain for \(f^{4}(x)+g^{4}(x)\)

Both \(f(x)^4\) and \(g(x)^4\) share the same restrictions: \(|x| \geq 1\) for \(f(x)\) and \(x eq 0\). Thus, the domain remains \((-\infty, -1] \cup [1, \infty)\), excluding \(x=0\).

Step 5: Find the formula for \((f \circ g)(x)\)

When finding \((f \circ g)(x)\), we substitute \(g(x)\) into \(f(x)\): \[(f \circ g)(x) = f(g(x)) = \sqrt{\left(\frac{2}{x}\right)^2 - 1} = \sqrt{\frac{4}{x^2} - 1}\].

Step 6: Determine the domain for \((f \circ g)(x)\)

\(g(x)=\frac{2}{x}\) requires \(x eq 0\) and must also satisfy \(\frac{4}{x^2} - 1 \geq 0\). This inequality implies \(\frac{4}{x^2} \geq 1\), thus \(|x| \leq 2\). The domain is \((-\infty, -2] \cup [2, \infty)\).

Step 7: Find the formula for \((g \circ f)(x)\)

To find \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\): \[(g \circ f)(x) = g(f(x)) = \frac{2}{\sqrt{x^2 - 1}}\].

Step 8: Determine the domain for \((g \circ f)(x)\)

The restriction \(f(x)\) gives \(|x| \geq 1\). To avoid division by zero in \(g(f(x))\), make sure \(\sqrt{x^2 - 1} eq 0\). Hence, domain of \((g \circ f)(x)\) is the same as \((-\infty, -1) \cup (1, \infty)\).

Key Concepts

Function Multiplication

Function multiplication involves multiplying two functions together to form a new function. In mathematical terms, if you have two functions, say \( f(x) \) and \( g(x) \), their multiplication is denoted as

• \((f \cdot g)(x) = f(x) \cdot g(x)\)

For instance, given \( f(x) = \sqrt{x^2 - 1} \) and \( g(x) = \frac{2}{x} \), their product is

• \((f \cdot g)(x) = \sqrt{x^2 - 1} \cdot \frac{2}{x} = \frac{2\sqrt{x^2 - 1}}{x}\)

This process sometimes requires you to simplify the expression to its cleanest form.
Multiplying functions requires attention to each function's characteristics, such as the presence of square roots or fractions, which can affect the expression's form and its domain.

Function Domain

The domain of a function is the set of input values (\(x\) values) for which the function is well-defined and produces real values as outputs. When working with composite functions, like multiplications or compositions, it’s essential to consider the domains of each participating function.For example, consider the function \( f(x) = \sqrt{x^2 - 1} \). The domain requires values of \( x \) that satisfy

• \(x^2 - 1 \geq 0\)

This means

• \(|x| \geq 1\), leading to the domain
  • • \((-\infty,-1] \cup [1,\infty)\)

To include \( g(x) = \frac{2}{x} \), it's also necessary that

• \(x eq 0\)

Thus, the domain for the function \((f \cdot g)(x)\) is further refined to exclude \( x = 0 \).
This comprehensive approach ensures that calculations involving these functions remain valid across their respective domains.

Function Composition

Function composition involves creating a new function by using one function as the input for another. If you have two functions \( f(x) \) and \( g(x) \), the composition \( f(g(x)) \) means that you plug \( g(x) \) into \( f(x) \). This is represented as:

• \((f \circ g)(x) = f(g(x))\)

For example, with \( f(x) = \sqrt{x^2 - 1} \) and \( g(x) = \frac{2}{x} \),

• \((f \circ g)(x) = \sqrt{\left(\frac{2}{x}\right)^2 - 1} \)

Function composition can also be represented the other way around, as \( g(f(x)) \), which means inserting \( f(x) \) into \( g(x) \):

• \((g \circ f)(x) = g(f(x)) = \frac{2}{\sqrt{x^2 - 1}}\)

Understanding these operations involves carefully managing each function's specific properties to ensure that they align correctly during the substitution process.

Square Root Functions

Square root functions are special mathematical functions involving the square root operation. They usually look like \( \sqrt{x} \) or in more complex forms, like \( \sqrt{x^2 - 1} \), as seen here. The construction \( f(x) = \sqrt{x^2 - 1} \) demands specific domain consideration, as the expression under the square root, or the radicand, must be non-negative:

• \(x^2 - 1 \geq 0\)

This constraint leads to a domain where the radicand is zero or positive. Additionally, square roots tend to simplify to absolute values, for example:

• \((\sqrt{x^2}) = |x| = x\) for \(x \geq 0\)

When working with square root functions, it’s crucial to understand how they limit the range of valid input values. Moreover, these functions introduce potential complications in expressions during operations like multiplication or composition, where simplification cannot always guarantee nonnegative expressions without careful analysis.