Problem 62 - Algebra and Trigonometry

Question

In Exercises \(61-64,\) find the domain of each function. $$ f(x)=\frac{1}{\sqrt{4 x^{2}-9 x+2}} $$

Step-by-Step Solution

Answer

The domain of the function \(f(x)= \frac{1}{\sqrt{4 x^{2}-9 x+2}}\) is \((-\infty, 0.5) \cup (1, \infty)\).

Step 1: Identify the given function

The given function is \(f(x)= \frac{1}{\sqrt{4 x^{2}-9 x+2}}\). This function can only be defined if the denominator is positive, as we cannot divide by zero and also cannot take the square root of a negative number.

Step 2: Find the domain

First, we need to solve the inequality \(4 x^{2} - 9 x + 2 > 0\). This is a quadratic inequality which can be solved by factorising the quadratic expression or by using the quadratic formula to find the x-values. This will give us the roots of the quadratic. Then, by evaluating values between the roots we can determine the intervals for which the expression under the square root is positive.

Step 3: Solve the quadratic inequality

Setting \(4 x^{2} - 9 x + 2 = 0\), this can be factored as \((2x - 1)(2x - 2) = 0\). Setting each factor equal to zero gives us the roots \(x = 0.5 , x = 1\). Now we need to check values of x between these roots, as well as to the left of 0.5 and to the right of 1. We find that \(4 x^{2} - 9 x + 2 > 0\) when \(x < 0.5\) or \(x > 1\). These are the values for which the function is defined. Thus, the domain of the function is \((-\infty, 0.5) \cup (1, \infty)\).

Key Concepts

Quadratic inequality

Quadratic inequalities are mathematical expressions that involve an inequality sign and a quadratic function. In our exercise, we aimed to find when the quadratic expression \(4x^2 - 9x + 2\) is positive, which consequently helps define the domain of the function \(f(x)=\frac{1}{\sqrt{4x^2-9x+2}}\). To solve a quadratic inequality, follow these steps:

• First, set the quadratic expression equal to zero to find the roots, using factorization or the quadratic formula. The roots, in this case, are \(x = 0.5\) and \(x = 1\).


• Next, the number line can be segmented into intervals based on these roots: \((-\infty, 0.5)\), \( (0.5, 1)\), and \((1, \infty)\).


• Test a value from each interval in the original inequality \(4x^2 - 9x + 2 > 0\) to see which intervals satisfy the inequality.

Typically, quadratic expressions will have different signs in alternate intervals. By testing values in these regions, you can determine for which intervals the inequality holds true. The solution to the inequality then represents the domain of the function.

Rational functions

Rational functions are quotients of two polynomials, such as \(f(x)= \frac{g(x)}{h(x)}\). The domain of a rational function is determined by identifying the x-values that do not cause division by zero. In our example, the function is \( f(x) = \frac{1}{\sqrt{4x^2 - 9x + 2}} \), which is a bit complex as it includes the square root in the denominator.

• The expression \( \sqrt{4x^2 - 9x + 2} \) must be greater than zero because a negative value under a square root is undefined and division by zero is not possible.


• Thus, the domain rule extends from finding values that ensure the denominator is a positive number, thus keeping the rational function defined.

Understanding this aspect of rational functions is crucial to solve domain-related problems effectively without falling into the division by zero pitfall.

Square root functions

Square root functions involve expressions under a radical symbol, requiring non-negative values underneath to be real and defined. In our problem, the square root in the denominator \( \sqrt{4x^2 - 9x + 2} \) dictates our approach to determining the domain of the function.

• To find the domain, we set up the condition that the expression under the square root must be greater than zero: \( 4x^2 - 9x + 2 > 0 \).


• This inequality ensures the expression remains non-negative, avoiding imaginary solutions and enabling real output values for the function.


• When the square root is in the denominator, it's critical that we never reach zero, as that would lead to undefined values.

By solving the inequality using testing intervals found through roots \(x = 0.5\) and \(x = 1\), we ensure all conditions are met, giving us the valid domain. Square root functions require careful analysis of their under-root expressions to maintain validity across their domains.