Problem 26 - A Graphical Approach to Precalculus with Limits

Question

For each pair of fimetions, (a) find ( \(f+g)(x),(f-g)(x),\) and \((f g)(x) ;\) (b) give the domains of the functions in part (a); (c) find \(\frac{t}{x}\) and give its domain; (d) find \(f \circ g\) and give is domain: and (e) find \(g \circ f\) and give its domain. Do not use a calculator. $$f(x)=|2 x-4|, g(x)=x+1$$

Step-by-Step Solution

Answer

The domains of all functions except \( \frac{f}{g} \) are \( \mathbb{R} \); the domain of \( \frac{f}{g} \) is \( \mathbb{R} \setminus \{-1\} \).

Step 1: Find the Sum of the Functions

To find \( (f+g)(x) \), add the expressions for \( f(x) \) and \( g(x) \): \[ (f+g)(x) = |2x-4| + (x+1) \]

Step 2: Find the Difference of the Functions

To find \( (f-g)(x) \), subtract the expression for \( g(x) \) from \( f(x) \):\[ (f-g)(x) = |2x-4| - (x+1) \]

Step 3: Find the Product of the Functions

To find \( (fg)(x) \), multiply the expressions for \( f(x) \) and \( g(x) \):\[ (fg)(x) = |2x-4|(x+1) \]

Step 4: Determine the Domain of (f+g)(x), (f-g)(x), and (fg)(x)

The domain for each of \( f(x) \) and \( g(x) \) is all real numbers, \( \mathbb{R} \), because there are no restrictions like division by zero or square roots of negative numbers. Hence, the domain of \( (f+g)(x) \), \( (f-g)(x) \), and \( (fg)(x) \) is also \( \mathbb{R} \).

Step 5: Find the Quotient of the Functions

To find \( \frac{f}{g} \), divide \( f(x) \) by \( g(x) \):\[ \frac{f}{g}(x) = \frac{|2x-4|}{x+1} \]

Step 6: Determine the Domain of the Quotient Function

The function \( \frac{f}{g}(x) = \frac{|2x-4|}{x+1} \) is defined for all real numbers except where the denominator is zero. Set \( x+1 = 0 \) to find the restriction: \( x = -1 \). Therefore, the domain is \( \mathbb{R} \setminus \{-1\} \).

Step 7: Find the Composition f(g(x))

To find \( f \circ g \), substitute \( g(x) \) into \( f(x) \):\[ f\circ g(x) = f(g(x)) = f(x+1) = |2(x+1) - 4| \] Simplify: \[ = |2x + 2 - 4| = |2x - 2| \]

Step 8: Determine the Domain of f(g(x))

Since \( g(x) \) and \( f(x) \) both have domains over all real numbers, the domain of \( f \circ g \) is also all real numbers, \( \mathbb{R} \).

Step 9: Find the Composition g(f(x))

To find \( g \circ f \), substitute \( f(x) \) into \( g(x) \):\[ g\circ f(x) = g(f(x)) = g(|2x-4|) = |2x-4| + 1 \]

Step 10: Determine the Domain of g(f(x))

For \( g \circ f \), the substitution does not introduce any new restrictions. Thus, the domain is \( \mathbb{R} \), as both \( f(x) \) and \( g(x) \) are defined for all real numbers.

Key Concepts

Domain of Functions

When working with functions, one of the key concepts to understand is the domain. The domain of a function is the set of all possible input values (usually 'x') that the function can accept without running into undefined scenarios, such as division by zero or taking the square root of a negative number.

• For simple functions such as linear functions, the domain is often all real numbers, denoted by \( \mathbb{R} \).
• To determine the domain, consider each operation in the function and identify any restrictions it may introduce.

In the case of our functions, \( f(x) = |2x-4| \) and \( g(x) = x+1 \), there are no inherent restrictions because we aren't dividing by a variable or calculating square roots. Thus, the domain of both \( f \) and \( g \) is \( \mathbb{R} \). Therefore, for combined functions like \((f+g)(x)\), \((f-g)(x)\), and \((fg)(x)\), the domain remains \( \mathbb{R} \). However, when considering the quotient \( \frac{f}{g}(x) \), avoid values that make the denominator zero. This occurs at \( x = -1 \), so the domain here is \( \mathbb{R} \setminus \{-1\} \).

Composition of Functions

Composition of functions involves applying one function to the results of another function. Notated as \( f \circ g \) or \( g \circ f \), it essentially means you substitute one function into another.

• To find \( f \circ g(x) \), plug \( g(x) \) into \( f(x) \).
• To find \( g \circ f(x) \), plug \( f(x) \) into \( g(x) \).

For our functions, we calculate:- \( f \circ g(x) = f(g(x)) = |2(x+1) - 4| = |2x - 2| \)- \( g \circ f(x) = g(f(x)) = |2x-4| + 1 \)The domains for these compositions depend on both component functions. Since both \( f(x) \) and \( g(x) \) are defined for all real \( x \), and substitution doesn't introduce new restrictions, both \( f \circ g \) and \( g \circ f \) inherit the domain \( \mathbb{R} \).

Absolute Value Functions

The absolute value function, denoted by \(|x|\), transforms any input into a non-negative output. It essentially measures the 'distance' of a number from zero on the number line.

• The expression \(|2x-4|\) implies the function takes \(2x-4\), determines its sign, and returns the non-negative value.
• Absolute value operations are essential for reflecting a function's behavior symmetrically about the y-axis when translated.

This property of symmetry means that absolute value functions have domains covering all real numbers, \( \mathbb{R} \), because no value of \( x \) makes the function undefined. Additionally, combining absolute functions with other operations affects their expression but not necessarily their domain. Thus, functions such as \( |2x-4| + (x+1) \) also have a domain of \( \mathbb{R} \).