Problem 15 - Precalculus with Limits

Question

In Exercises 9-16, find (a) \((f+g)(x)\), (b) \((f-g)(x)\), (c) \((fg)(x)\), and (d) \((f/g)(x). What is the domain of \)f/g\(? \)f(x) = \frac{1}{x}\(, \)g(x) = \frac{1}{x^2}$

Step-by-Step Solution

Answer

\((f+g)(x) = \frac{x + 1}{x^2}\), \((f-g)(x) = \frac{x - 1}{x^2}\), \((fg)(x) = \frac{1}{x^3}\), \((f/g)(x) = x\) and the domain of \(f/g\) is all real numbers.

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

Now, add \(f(x)\) and \(g(x)\) together. \[ (f+g)(x) = f(x) + g(x) = \frac{1}{x} + \frac{1}{x^2} = \frac{x + 1}{x^2} \]

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

Then, subtract \(g(x)\) from \(f(x)\). \[ (f-g)(x) = f(x) - g(x) = \frac{1}{x} - \frac{1}{x^2} = \frac{x - 1}{x^2} \]

Step 3: Find (fg)(x)

Multiply \(f(x)\) with \(g(x)\). \[ (fg)(x) = f(x) \cdot g(x) = \frac{1}{x} \cdot \frac{1}{x^2} = \frac{1}{x^3} \]

Step 4: Find (f/g)(x)

Now, divide \(f(x)\) by \(g(x)\).\[ (f/g)(x) = \frac{f(x)}{g(x)} = \frac{1/x}{1/x^2} = x \]

Step 5: Find the domain of f/g

The domain of \(f/g\) is all real numbers except for those values that make the denominator zero. Since there's no denominator in \(f/g\), the domain of \(f/g\) is all real numbers.

Key Concepts

Domain of a Function

The domain of a function represents all the values that can be input into the function without causing any issues like division by zero or square roots of negative numbers. It's like the set of all possible inputs for the function. Each type of function has different rules for its domain.
For rational functions, the inputs (x-values) making the denominator zero are excluded from the domain, because division by zero is undefined. This is crucial when dealing with functions like \( f(x) = \frac{1}{x} \).

• If the function is \( f(x) = \frac{1}{x}\), then \( x eq 0\) because \( \frac{1}{0} \) is undefined.
• For \( g(x) = \frac{1}{x^2}\), it also holds that \( x eq 0\).

When considering operations on functions, especially division, the domain is determined by both functions involved. For division like \( (f/g)(x) = \frac{f(x)}{g(x)} \), the domain will exclude values making either \( g(x) \) zero or those already excluded from \( f(x) \). In our problem, since \( (f/g)(x) = x \) simplifies to no denominators, the domain could initially seem to be all real numbers, except origination constraints still apply.

Addition of Functions

Adding functions means that you take two functions and add their outputs together. This leads to a new function whose domain is the intersection of the domains from the added functions.
In math terms, if \( f(x) \) and \( g(x) \) are two functions, the addition is defined as \((f+g)(x) = f(x) + g(x)\).
This operation means you perform the arithmetic addition of \( f(x) \) and \( g(x) \), ensuring you only input values (x) belonging to both function domains.

• For our specific exercise:\[ (f+g)(x) = \frac{1}{x} + \frac{1}{x^2} = \frac{x + 1}{x^2} \]

Keep in mind that the domain of \((f+g)\) remains \( x eq 0 \), as this is the overlapping restriction from both functions.

Subtraction of Functions

The subtraction of functions is very similar to addition in terms of domain and execution, but instead of adding the outputs of two functions, you subtract them. Subtraction between functions \( f(x) \) and \( g(x) \) is given by \((f-g)(x) = f(x) - g(x)\).
Just like addition, you use only x-values that fit within both functions' domains.

• In the exercise example, the subtraction becomes:\[ (f-g)(x) = \frac{1}{x} - \frac{1}{x^2} = \frac{x - 1}{x^2} \]

The domain constraint remains as before: \( x eq 0 \). Remember, the common domain (intersection) is necessary for subtraction.

Multiplication of Functions

When multiplying functions, you take each function's output and multiply them. The domain of the product of two functions is also the intersection of their domains, where both operate without restrictions.
Mathematically, for functions \( f(x) \) and \( g(x) \), the multiplication is defined as \((fg)(x) = f(x) \cdot g(x)\).
This approach ensures input x-values are valid for both functions separately.

• For our exercise's multiplication:\[ (fg)(x) = \frac{1}{x} \cdot \frac{1}{x^2} = \frac{1}{x^3} \]

Again, don't forget that the domain \( x eq 0 \) applies in this case due to the denominator.

Division of Functions

Dividing functions can be trickier due to the additional requirement of not letting the denominator become zero when forming the quotient. The rule for division is: for functions \( f(x) \) and \( g(x) \), division is formulated as \((f/g)(x) = \frac{f(x)}{g(x)} \), where \( g(x) eq 0 \).
Besides excluding values from both individual domains, the division also excludes values making \( g(x) = 0 \).

• In the specific problem exercise:\[ (f/g)(x) = \frac{\frac{1}{x}}{\frac{1}{x^2}} = x \]
• Note the function simplifies crisply to \( x \), with no denominator, but as per individual constraints usually excluded \( x = 0 \).

In essence, reflect back to initial expressions of both functions to confirm domain collective rules.