Problem 61 - Algebra and Trigonometry

Question

In Exercises \(61-64,\) find the domain of each function. $$ f(x)=\sqrt{2 x^{2}-5 x+2} $$

Step-by-Step Solution

Answer

The domain of the function \(f(x) = \sqrt{2x^2 - 5x + 2}\) is \(1 \leq x \leq 2\).

Step 1: Write Down the Function

We begin with the function \(f(x) = \sqrt{2x^2 - 5x + 2}\). We are looking for the domain of this function, or in other words, the set of all x-values for which this function is defined.

Step 2: Understand Under What Conditions the Function is Undefined

The function will be undefined where the expression inside the square root is negative. This is because square roots of negative numbers are not defined in the real number system.

Step 3: Set the Inside of the Square Root Greater Than or Equal to Zero

In order to find the domain we have to find the values of x for which the inside of the square root, \(2x^2 - 5x + 2\), is greater than or equal to zero. Therefore, we need to solve the inequality \(2x^2 - 5x + 2 \geq 0\).

Step 4: Solve the Quadratic Inequality

First we factor the quadratic expression. It factors to \(2(x - 1)(x - 2) \geq 0\). Now we set each factor individually equal to zero to find the critical points which are the borders of the intervals we will later test. We have \(x - 1 = 0\) and \(x - 2 = 0\), so the critical points are \(x = 1\) and \(x = 2\).

Step 5: Test the Intervals

The critical points \(x = 1\) and \(x = 2\) divide the number line into three intervals: \(-\infty, 1)\), \(1, 2)\), and \(2, +\infty\). We choose a test point in each interval and substitute it into the inequality. Any value less than 1 can be used for the first interval, so let’s use \(x = 0\). Substituting into the inequality gives \(2(0 - 1)(0 - 2) \geq 0\) which is false, so the interval \(-\infty, 1)\) is not part of the solution. For the second interval, we take a point between 1 and 2, let’s use \(x = 1.5\). Substituting into the inequality gives \(2(1.5 - 1)(1.5 - 2) \geq 0\) which is true, so the interval \(1,2]\) is part of the solution. For the third interval, we can take any point greater than 2, let’s use \(x = 3\). Substituting into the inequality we see that the inequality is false, hence \(2, +\infty\) is not part of the solution.

Step 6: Write Down the Solution

So the solution to the inequality \(2x^2 - 5x + 2 \geq 0\) is \(1 \leq x \leq 2\). Thus, the domain of the function \(f(x) = \sqrt{2x^2 - 5x + 2}\) is \(1 \leq x \leq 2\).

Key Concepts

Quadratic Inequality

A quadratic inequality involves an expression like a quadratic equation, but instead of an equal sign, you have an inequality symbol such as ">", "<", "≥", or "≤". In our function, the inequality is at the heart of finding the domain. We solved the inequality \(2x^2 - 5x + 2 \geq 0\).

This inequality tells us where the expression within the square root is valid (i.e., where it is non-negative). We must solve for \(x\) to find the ranges that satisfy this condition.

The solutions to these inequalities give us intervals on the number line, which are crucial to determining the valid input values for the function. These valid intervals represent the domain of the function itself. For our given problem, solving a quadratic inequality is imperative to find when the function remains real and defined.

Critical Points

Critical points are values of \(x\) at which the expression inside a function changes its behavior. In the case of quadratic inequalities, these appear when the quadratic expression equals zero.

For \(2x^2 - 5x + 2\), setting the expression to zero gives us the points \(x = 1\) and \(x = 2\). These points are important because they partition the number line into segments.

Once the critical points are identified, they help determine the intervals on the number line to test for validity in inequalities. Essentially, they mark boundaries of regions where the inequality might hold true.

Using critical points, one tests each section to confirm where the inequality satisfies the conditions set by the problem, thus aiding in outlining the domain.

Factoring

Factoring is an essential technique for solving quadratic inequalities. In our example, we factored the expression into \(2(x - 1)(x - 2)\). This made solving the inequality \(2x^2 - 5x + 2 \geq 0\) easier, as it broke down the quadratic into approachable linear components.

Factoring transforms a complex quadratic expression into a product of simpler ones that are easier to work with. This allows us to quickly find the critical points and understand the inequality's behavior across different intervals.

By finding roots and testing intervals, factoring provides a structured path to solving the inequality and determining where the expression is positive, aiding significantly in finding the real domain.

Real Number System

The real number system encompasses all measurable values, including both rational and irrational numbers. In the context of this exercise, it's important because we're concerned with values that keep our function within this system.

Why? Because the square root function is not defined for negative numbers within the real number system. Therefore, ensuring the expression within the square root \(2x^2 - 5x + 2\) is non-negative restricts the domain to only those x-values.

This need to maintain the function within the realm of real numbers dictates the strategy for finding its domain. Ultimately, by confining x-values to those resulting in real and define expressions, we ensure our function remains valid over its defined domain.