Problem 7 - Algebra for College Students

Question

\(f(x)=\sqrt{x-1}, \quad g(x)=\sqrt{x}\)

Step-by-Step Solution

Answer

Both compositions are defined for \( x \geq 1 \).

Step 1: Understanding the Functions

We have two functions: \( f(x) = \sqrt{x-1} \) and \( g(x) = \sqrt{x} \). The function \( f(x) \) is defined for \( x \geq 1 \) because the expression inside the square root must be non-negative. Similarly, \( g(x) \) is defined for \( x \geq 0 \). Both functions involve square roots, so we need to ensure we only consider values of \( x \) within their respective domains.

Step 2: Solving for f(x)

The function \( f(x) = \sqrt{x-1} \) transforms a number by subtracting 1 from \( x \) and then taking the square root of the result. For \( f(x) \) to be real, \( x-1 \geq 0 \), which means \( x \geq 1 \).

Step 3: Solving for g(x)

The function \( g(x) = \sqrt{x} \) takes the square root of \( x \). For \( g(x) \) to be real, \( x \geq 0 \).

Step 4: Finding the Composition of f(g(x))

To find \( f(g(x)) \), substitute \( g(x) \) into \( f(x) \): \( f(g(x)) = f(\sqrt{x}) = \sqrt{\sqrt{x} - 1} \). For this expression to be real, \( \sqrt{x} - 1 \geq 0 \), which means \( \sqrt{x} \geq 1 \). Solving \( \sqrt{x} \geq 1 \) gives \( x \geq 1 \). Therefore, \( x \) must be greater than or equal to 1 to make \( f(g(x)) \) valid.

Step 5: Finding the Composition of g(f(x))

To find \( g(f(x)) \), substitute \( f(x) \) into \( g(x) \): \( g(f(x)) = g(\sqrt{x-1}) = \sqrt{\sqrt{x-1}} \). For this expression to be real (within the bounds of \( g(x) \)), \( x-1 \geq 0 \) must hold: \( x \geq 1 \). Thus, \( g(f(x)) \) is defined for \( x \geq 1 \).

Key Concepts

Square Root Functions

A square root function takes the form \(f(x) = \sqrt{x}\). It involves finding a number which, when multiplied by itself, equals \(x\). Here, square root functions are used in two instances:

• \(f(x) = \sqrt{x - 1}\)
• \(g(x) = \sqrt{x}\)

For both functions, the symbol \(\sqrt{}\) is crucial. It indicates that each input value after performing any necessary operations will be transformed by finding its square root. This function is prevalent in mathematical models that deal with quadratic relationships or in scenarios requiring balancing or reducing numerical values. Keep in mind, the output of a square root function is always non-negative, as it provides the principal square root.

Function Domain

The domain of a function refers to the set of all possible input values \(x\) that the function can accept without resulting in undefined or non-real values. For functions involving square roots, the expression inside the root must be non-negative:

• For \(f(x) = \sqrt{x-1}\), the domain is determined by \(x-1 \geq 0\). Hence, \(x \geq 1\).
• For \(g(x) = \sqrt{x}\), we consider \(x \geq 0\) as the domain.

It is crucial to evaluate the domain before performing function operations, like composition, to ensure that the resultant functions remain valid across the intended scope of input values.

Real-Valued Functions

Real-valued functions are those where the range, or set of possible output values, consists of real numbers. Working with square root functions implicates that our resultant values must also adhere to this principle of yielding real numbers only:

• For \(f(x) = \sqrt{x-1}\), the condition \(x \geq 1\) ensures that all outputs are real.
• For \(g(x) = \sqrt{x}\), by setting \(x \geq 0\), the function is designed to yield real number outcomes.

When composing functions like \(f(g(x))\) or \(g(f(x))\), maintaining real-valued outputs means considering the overlap of both functions' domains. This ensures that every input value results in a real and non-imaginary output, conserving the integrity of real-valued functions in mathematical computations.