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Write the domain of the function in interval notation. $$ q(x)=\sqrt{4 x^{2}+7 x-2} $$

Short Answer

Expert verified
The domain is \((-\

Step by step solution

01

Identify the domain condition for the square root function

For the function under a square root to be defined, the expression inside the square root must be non-negative. Therefore, we need to find the values of x for which the expression inside the square root, \(4x^2 + 7x - 2\), is greater than or equal to 0.
02

Find the zeros of the quadratic equation

To find the values of x where \(4x^2 + 7x - 2 = 0\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 7\), and \(c = -2\). Calculate the discriminant first: \(b^2 - 4ac = 7^2 - 4(4)(-2) = 49 + 32 = 81\). Then the roots are \(x = \frac{-7 \pm \sqrt{81}}{2(4)} = \frac{-7 \pm 9}{8}\).
03

Find the critical points

Solving for x, we get two roots: \(x = \frac{-7 + 9}{8} = \frac{1}{4}\) and \(x = \frac{-7 - 9}{8} = -2\).
04

Determine the intervals to test

The quadratic opens upwards (since \(a = 4 > 0\)), and the critical points divide the number line into three intervals: \((-\infty, -2)\), \((-2, \frac{1}{4})\), and \(\left(\frac{1}{4}, +\infty\right)\). We will test points within each interval to determine where the quadratic is non-negative.
05

Test the intervals

Choose test points: For \((-\infty, -2)\), let's pick \(x = -3\). For \((-2, \frac{1}{4})\), let's pick \(x = 0\). For \(\left(\frac{1}{4}, +\infty\right)\), let's pick \(x = 1\). Calculate \(4x^2 + 7x - 2\) for each test point to see if it is non-negative.
06

Evaluate the quadratic at test points

Evaluate: \(4(-3)^2 + 7(-3) - 2 = 36 - 21 - 2 = 13\) which is positive; \(4(0)^2 + 7(0) - 2 = -2\) which is negative; \(4(1)^2 + 7(1) - 2 = 4 + 7 - 2 = 9\) which is positive.
07

Combine the intervals

The quadratic is non-negative in the intervals where the test points evaluation is positive. Therefore, the intervals where \(4x^2 + 7x - 2 \geq 0\) are \((-\infty, -2] \cup [\frac{1}{4}, +\infty)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equation
A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(a, b\), and \(c\) are constants and \(a eq 0\). Quadratic equations are fundamental in algebra and appear in many fields of science. The general solution to a quadratic equation is given by the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. This formula helps us find the values of \(x\) where the quadratic equation equals zero, known as the roots or solutions of the equation.

In our exercise, the quadratic equation \(4x^2 + 7x - 2\) is found inside the square root function. To find where this function is defined, we need to solve for the values of \(x\) such that the expression under the square root is non-negative (i.e., \(4x^2 + 7x - 2 \geq 0\)). This is key for determining the domain of the function.

Interval Notation
Interval notation is a way of writing subsets of the real number line. It uses parentheses and brackets to describe the set of numbers between two endpoints. Parentheses \(()\) denote open intervals where the endpoints are not included, while brackets \([]\) indicate closed intervals where the endpoints are included.

For example, \((-2, 5)\) represents all numbers between \(-2\) and \(5\), not including \(-2\) and \(5\) themselves. Conversely, \([-2, 5]\) includes both \(-2\) and \(5\).

In the context of our problem, once we find the roots of the quadratic equation, we use interval notation to describe the domain of the function. Here we determined that \(4x^2 + 7x - 2\) is non-negative in the intervals \((-\infty, -2]\) and \([\frac{1}{4}, +\infty)\). These intervals describe where the function \(q(x)=\sqrt{4x^2+7x-2}\) is defined and hence forms its domain.
Discriminant
The discriminant is a part of the quadratic formula and is crucial in understanding the nature of the roots of a quadratic equation. It is given by the expression \(b^2 - 4ac\).

The discriminant can tell us the following:

  • If \(b^2 - 4ac > 0\), the quadratic equation has two distinct real roots.

  • If \(b^2 - 4ac = 0\), the quadratic equation has exactly one real root (also called a double root).

  • If \(b^2 - 4ac < 0\), the quadratic equation has no real roots but two complex roots.


For our quadratic equation \(4x^2 + 7x - 2\), we calculated the discriminant as follows:
\(b^2 - 4ac = 7^2 - 4(4)(-2) = 49 + 32 = 81\). Since the discriminant is positive (\(81 > 0\)), we know that the quadratic equation has two distinct real roots, \(-2\) and \(\frac{1}{4}\). These roots are critical in determining the intervals of our domain.

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