Problem 30 - A Graphical Approach to College Algebra

Question

For each pair of functions, (a) find \((f+g)(x),(f-g)(x),\) and \((f g)(x)\) (b) give the domains of the functions in part (a); (c) find \(\frac{f}{8}\) and give its domain; (d) find \(f \circ g\) and give its domain; and (e) find \(g \circ f\) and give its domain. Do not use a calculator. $$f(x)=\sqrt{2+4 x^{2}}, g(x)=x$$

Step-by-Step Solution

Answer

The domains for all functions are all real numbers, (- infty, infty).

Step 1: Find (f+g)(x)

To find \((f+g)(x)\), simply add the two given functions. We have:\[(f+g)(x) = f(x) + g(x) = \sqrt{2+4x^2} + x\]

Step 2: Find (f-g)(x)

For \((f-g)(x)\), subtract the second function from the first. Thus:\[(f-g)(x) = f(x) - g(x) = \sqrt{2+4x^2} - x\]

Step 3: Find (f*g)(x)

Multiply the two functions together to obtain \((f*g)(x)\):\[(fg)(x) = f(x) \cdot g(x) = \sqrt{2+4x^2} \cdot x\]

Step 4: Determine the domain of (f+g)(x), (f-g)(x), and (f*g)(x)

The function \(f(x) = \sqrt{2+4x^2}\) requires the expression under the square root to be non-negative, hence \(2 + 4x^2 \geq 0\), which is always true. Therefore, the domain of \(f(x)\) is all real numbers, and similarly the domain of \(g(x) = x\) is all real numbers. Hence, the domain of \((f+g)(x), (f-g)(x), (f*g)(x)\) is all real numbers \((-infty, infty)\).

Step 5: Find f/8 and its domain

The expression \( \frac{f}{8} \) is:\[\frac{f(x)}{8} = \frac{\sqrt{2+4x^2}}{8}\]Since no division by zero occurs, the domain is the same as that of \(f(x)\), which is all real numbers \((-infty, infty)\).

Step 6: Find the composition f ∘ g and its domain

To find \(f \circ g\), substitute \(g(x)\) into \(f(x)\): \[(f \circ g)(x) = f(g(x)) = f(x) = \sqrt{2+4x^2}\]The domain is that of \(f(x)\), which is all real numbers \((-infty, infty)\).

Step 7: Find the composition g ∘ f and its domain

To find \(g \circ f\), substitute \(f(x)\) into \(g(x)\):\[(g \circ f)(x) = g(f(x)) = f(x) = \sqrt{2+4x^2}\]The domain is also all real numbers \((-infty, infty)\).

Key Concepts

Function Operations

Function operations let us combine two functions in various ways such as addition, subtraction, multiplication, and division. These basic operations are very similar to operations you perform on numbers.

For example, when working with functions like \( f(x) = \sqrt{2+4x^2} \) and \( g(x) = x \), you can explore:

• Adding them: \( (f+g)(x) = \sqrt{2+4x^2} + x \). This means you simply add \( f(x) \) and \( g(x) \) at each value of \( x \).
• Subtracting them: \( (f-g)(x) = \sqrt{2+4x^2} - x \). This involves taking \( g(x) \) away from \( f(x) \).
• Multiplying them: \( (f\cdot g)(x) = \sqrt{2+4x^2} \cdot x \). Here you multiply \( f(x) \) and \( g(x) \).
• Even dividing by a constant, like \(\frac{f(x)}{8}\), resulting in \({\frac{\sqrt{2+4x^2}}{8}} \).

Understanding how these operations work lays the foundation for more complex function manipulations.

Domain of Functions

The domain of a function is all the possible input values (\(x\)-values) that the function can accept without causing any mathematical errors. For a function like \(f(x) = \sqrt{2+4x^2}\), the expression inside the square root, \(2+4x^2\), must be non-negative because a square root of a negative number is not real.

However, \(2+4x^2\) is non-negative for all real \(x\), so \(f(x)\) is defined for every real number. This makes the domain \((-\infty, \infty)\). Similarly, since \(g(x) = x\) is defined for all real numbers, its domain is also \((-\infty, \infty)\).

• Thus, the domain of both \(f+g\) and \(f-g\) is \((-\infty, \infty)\).
• The division by 8 in \(\frac{f(x)}{8}\) doesn’t affect the domain, so it also remains \((-\infty, \infty)\).

Understanding domains is crucial to ensure functions are calculated correctly and to avoid undefined operations.

Composition of Functions

Composition of functions involves plugging one function into another. This is written as \(f \circ g\) or \(g \circ f\). Here’s how it works:

When you perform \(f \circ g\), you substitute \(g(x)\) into \(f(x)\). For instance, since \(g(x) = x\), substituting into \(f(x)\) gives \( (f \circ g)(x) = f(g(x)) = \sqrt{2+4x^2}\). Thus, the result is \(f(x)\) itself.

Similarly, \(g \circ f\) means substituting \(f(x)\) into \(g(x)\). If \(g(x)\) is \(x\), then \(g \circ f\) results in \(\sqrt{2+4x^2}\), the same as \(f(x)\). Full understanding of function composition is key for advanced mathematical studies such as calculus because it helps in creating new functions and solving complex equations.