Problem 15 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find (a) \((f+g)(x),\) (b) \((f-g)(x)\) , (c) \((f g)(x),\) and \((d)(f / g)(x) .\) What is the domain of \(f / g ?\) $$f(x)=x^{2}+5, \quad g(x)=\sqrt{1-x}$$

Step-by-Step Solution

Answer

The results are: (a) \((f+g)(x) = x^2 + 5 + \sqrt{1-x}\), (b) \((f-g)(x) = x^2 + 5 - \sqrt{1-x}\), (c) \((f g)(x) = (x^2 + 5) . \sqrt{1-x}\), (d) \((f / g)(x) = (x^2 + 5) / \sqrt{1-x}\), and the domain of the quotient \(f/g\) is \(-\infty,1)\.

Step 1: Addition of Functions

The addition is done by simply adding the rule of each function: \((f+g)(x) = f(x) + g(x) = x^2 + 5 + \sqrt{1-x}\)

Step 2: Subtraction of Functions

Subtraction can be done likewise by subtracting the rule of g(x) from f(x): \((f-g)(x) = f(x) - g(x) = x^2 + 5 - \sqrt{1-x}\)

Step 3: Multiplication of Functions

The rule for the product of f(x) and g(x) is determined by multiplying their individual rules: \((f g)(x) = f(x) . g(x) = (x^2 + 5) . \sqrt{1-x}\)

Step 4: Division of Functions

The rule for the quotient of f(x) and g(x) is obtained by dividing the rule of f(x) by the rule of g(x): \((f / g)(x) = f(x) / g(x) = (x^2 + 5) / \sqrt{1-x}\)

Step 5: Find the domain of f/g

The domain of \(f / g\), or \(f(x) / g(x)\), will be all real numbers except where \(g(x) = 0 \). Since \(g(x) = \sqrt{1-x}\), the domain is all the real values of x such that \(1-x > 0\) or \(x < 1\). So, the domain of \(f/g\) is \(-\infty,1)\.

Key Concepts

Addition of Functions

In mathematics, adding functions is a straightforward process. To find the sum of two functions, you simply add the output (or value) of each function for a given input. If you have two functions, say \(f(x)\) and \(g(x)\), the addition of these functions is represented as \((f+g)(x) = f(x) + g(x)\). This means you take the expression for \(f(x)\) and add it to the expression for \(g(x)\).

In our example:

• \(f(x) = x^2 + 5\)
• \(g(x) = \sqrt{1-x}\)

The function formed by adding them will be \((f+g)(x) = x^2 + 5 + \sqrt{1-x}\). This new function \((f+g)(x)\) gives you the sum of the values of both functions for any input \(x\). It's essential to pay attention to operations to ensure proper addition.

Subtraction of Functions

Subtracting one function from another is as simple as it sounds. For functions \(f(x)\) and \(g(x)\), the subtraction is expressed as \((f-g)(x) = f(x) - g(x)\). This involves taking each corresponding output of \(f(x)\) and \(g(x)\) and subtracting them.

For instance, with our functions:

• \(f(x) = x^2 + 5\)
• \(g(x) = \sqrt{1-x}\)

The subtraction would result in \((f-g)(x) = x^2 + 5 - \sqrt{1-x}\). This operation generates a new function, which represents the difference in output values of \(f(x)\) and \(g(x)\) for any input \(x\). The important part is ensuring you maintain the correct order of subtraction.

Multiplication of Functions

When multiplying functions, the output at any value of \(x\) is found by multiplying the outputs of each function. If \(f(x)\) and \(g(x)\) are your two functions, multiplication is denoted by \((fg)(x) = f(x) \cdot g(x)\). This involves taking the expression for \(f(x)\) and multiplying it by \(g(x)\).

For these functions:

• \(f(x) = x^2 + 5\)
• \(g(x) = \sqrt{1-x}\)

The product is \((fg)(x) = (x^2 + 5) \cdot \sqrt{1-x}\). This function represents a combination where you multiply the rules of \(f\) and \(g\) for each input \(x\). Multiplication may lead to more complex expressions, so it's crucial to be comfortable with algebraic manipulation.

Division of Functions

Dividing functions is a bit different because it involves ensuring that the denominator does not equal zero. The division of two functions \(f(x)\) by \(g(x)\) is represented as \((f/g)(x) = \frac{f(x)}{g(x)}\).

For the given functions:

• \(f(x) = x^2 + 5\)
• \(g(x) = \sqrt{1-x}\)

The quotient is expressed as \((f/g)(x) = \frac{x^2 + 5}{\sqrt{1-x}}\). Since division by zero is undefined, you must ensure that \(g(x) eq 0\). This requires considering the domain to find values of \(x\) where the denominator remains positive.

Domain of a Function

Understanding the domain of a function is critical, especially for operations like division involving functions. The domain is the set of all possible input values (\(x\)) for which the function is defined.

Consider \(f/g\):

• \(f(x) = x^2 + 5\)
• \(g(x) = \sqrt{1-x}\)

The function \(g(x)\) involves a square root, meaning its domain is limited to where the expression inside remains non-negative, i.e., \(1-x > 0\), which simplifies to \(x < 1\). This results in a domain for \(f/g\) of \(-\infty, 1\).

It is crucial to correctly identify the domain to avoid undefined values and ensure the function's result is valid for every input.