Problem 27 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-32: Find the domain of \(f\). $$ f(x)=\frac{\sqrt{2 x-3}}{x^{2}-5 x+4} $$

Step-by-Step Solution

Answer

The domain is \(\left[\frac{3}{2},1\right) \cup (1,4) \cup (4,\infty)\)."

Step 1: Determine the domain restrictions from the numerator

The numerator is \(\sqrt{2x-3}\). For this square root to be defined, the expression inside must be greater than or equal to zero. Set up the inequality: \[2x-3 \geq 0\]Solve for \(x\):\[2x \geq 3\]\[x \geq \frac{3}{2}\]

Step 2: Determine the domain restrictions from the denominator

The denominator is \(x^2-5x+4\). For the function to be defined, the denominator must not be zero. Set the denominator equal to zero and solve for \(x\):\[x^2-5x+4=0\]Factor the quadratic expression:\[(x-1)(x-4)=0\]The solutions are\[x=1 \quad \text{and} \quad x=4\]These are the values that make the denominator zero, so they must be excluded from the domain.

Step 3: Combine the restrictions from numerator and denominator

The domain of the function \(f(x)=\frac{\sqrt{2x-3}}{x^2-5x+4}\) is determined by combining the restrictions found in Steps 1 and 2. \(- \) From Step 1, \(x\) must be greater than or equal to \(\frac{3}{2}\) for the numerator's square root to be defined.\(- \) From Step 2, \(x\) cannot be \(1\) or \(4\) because they make the denominator zero.Therefore, the domain is all \(x\) such that \(x \geq \frac{3}{2}\), but \(x eq 1\) and \(x eq 4\). This can be expressed in interval notation as:\[\left[\frac{3}{2},1\right) \cup (1,4) \cup (4,\infty)\]

Key Concepts

Square Root Function

A square root function involves finding the value that, when multiplied by itself, yields the original number inside the square root. In the function provided, we see a square root in the numerator: \( \sqrt{2x-3} \). For a square root function to be defined — that is, to produce real numbers — the expression under the square root symbol must be non-negative.

• This means you solve the inequality \( 2x - 3 \geq 0 \) to determine valid input values for \( x \).
• Solving this inequality gives \( x \geq \frac{3}{2} \), which is necessary for the square root to be "well-behaved" and to avoid dealing with complex numbers.

Starting with the correct conditions for a square root function ensures that any further calculations are based on real, not complex, numbers.

Quadratic Expression

Quadratic expressions are polynomial expressions of the form \( ax^2 + bx + c \). In the denominator of the given function, we have the quadratic expression \( x^2 - 5x + 4 \). Identifying where this expression equals zero is crucial, as these values would make the function undefined.
To deal with this quadratic expression:

• We factor \( x^2 - 5x + 4 \) to find its roots: \((x-1)(x-4)\).
• Setting each factor equal to zero, the solutions are \( x = 1 \) and \( x = 4 \). These are the points where the expression becomes zero, hence they cause the function to be undefined due to division by zero.

Understanding how to manage quadratic expressions is key to finding the domain when they appear in function denominators.

Inequality Solving

Solving inequalities is a fundamental skill used in determining whether a domain is valid for a given function. Inequalities help establish which values can be plugged into the function without causing issues like division by zero or taking the square root of a negative number.
In the context of finding a domain:

• You need to handle inequalities from both the numerator and denominator restrictions. From the numerator, \( 2x - 3 \geq 0 \) was solved to find \( x \geq \frac{3}{2} \).
• For the quadratic expression in the denominator, factor the equation and determine the critical point values that make the expression equal to zero.

This process of solving inequalities highlights which value ranges are permissible and which must be excluded, forming a coherent understanding of a function's domain.

Interval Notation

Interval notation is a concise way of expressing a range of values, especially useful when explaining the domain of a function. It clearly indicates the start and end of a range and specifies whether these endpoints are included or excluded.
For the function \( f(x) = \frac{\sqrt{2x-3}}{x^2-5x+4} \):

• The valid domain was determined to be all \( x \) such that \( x \geq \frac{3}{2} \) while excluding \( x = 1 \) and \( x = 4 \).
• This is notated as \[ \left[ \frac{3}{2}, 1 \right) \cup (1, 4) \cup (4, \infty) \], where brackets \( [ ] \) mean the endpoint is included, and parentheses \( ( ) \) mean it is not.

Mastering interval notation is essential for communicating a function’s domain accurately and succinctly.