Problem 41 - Precalculus Mathematics for Calculus

Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$2 x^{2}+x \geq 1$$

Step-by-Step Solution

Answer

The solution is \((-\infty, -1] \cup [\frac{1}{2}, \infty)\).

Step 1: Set the Inequality to Zero

Start by moving all the terms to one side of the inequality to set it to zero. We have: \[ 2x^2 + x - 1 \geq 0 \]

Step 2: Factor the Quadratic Expression

Factor the quadratic expression on the left side. We need two numbers that multiply to \(-2\) (the product of 2 and -1) and add to \(+1\) (the coefficient of \(x\)). These numbers are 2 and -1. This gives the factorization: \[ (2x - 1)(x + 1) \geq 0 \]

Step 3: Determine the Critical Points

The critical points arise from setting each factor equal to zero. Solve for \(x\): - From \(2x - 1 = 0\), we get \(x = \frac{1}{2}\). - From \(x + 1 = 0\), we get \(x = -1\). These critical points divide the number line into intervals.

Step 4: Test the Intervals

Determine which intervals satisfy the inequality by testing a sample point from each interval: 1. \((-\infty, -1)\): Try \(x = -2\). \((2(-2) - 1)(-2 + 1) = (-5)(-1) = 5 > 0\). 2. \((-1, \frac{1}{2})\): Try \(x = 0\). \((2(0) - 1)(0 + 1) = (-1)(1) = -1 ot\geq 0\). 3. \((\frac{1}{2}, \infty)\): Try \(x = 1\). \((2(1) - 1)(1 + 1) = (1)(2) = 2 > 0\). The inequality holds for intervals \((-\infty, -1)\) and \((\frac{1}{2}, \infty)\).

Step 5: Include Critical Points

The inequality is \(\geq\) so include the critical points where the expression equals zero. Test these critical points:- At \(x = -1\), \((2(-1) - 1)(-1 + 1) = (-3)(0) = 0\geq 0\). - At \(x = \frac{1}{2}\), \((2(\frac{1}{2}) - 1)(\frac{1}{2} + 1) = (0)(\frac{3}{2}) = 0 \geq 0\). Both points satisfy the inequality.

Step 6: Write the Solution in Interval Notation

Combine results from the tests. Since \([-1, -1]\) and \(\left[\frac{1}{2}, \frac{1}{2}\right]\) are included and valid: The solution in interval notation is: \[ (-\infty, -1] \cup [\frac{1}{2}, \infty) \]

Step 7: Graph the Solution Set

Graph the solution on the number line:- Use a filled circle at \(-1\) and \(\frac{1}{2}\) to indicate these points are included.- Draw a solid line extending from \(-\infty\) to \(-1\) and from \(\frac{1}{2}\) to \(\infty\).

Key Concepts

Interval Notation

Interval notation is a way of representing subsets of the real number line. It uses intervals to show where solutions to inequalities lie.
When solving inequalities, the solution is often a range of numbers that satisfy the condition.
Here’s how interval notation works:

• Use parentheses, \((\) and \(),\) to indicate that an endpoint is not included in the interval.
• Use brackets, \([\) and \(),\) to include an endpoint in the interval.
• For example, \((-\infty, -1]\) means all numbers less than or equal to -1. On the other side, \([\frac{1}{2}, \infty)\) includes all numbers greater than or equal to \frac{1}{2}\.
• The symbol \(\infty\) (infinity) indicates that there is no bound in that direction.

Interval notation is concise and well-suited for expressing solutions to inequalities, especially when combined with unions (\(\cup\)) to represent multiple intervals.

Quadratic Expression

A quadratic expression is a polynomial where the highest power of the variable is 2, typically written in the form \(ax^2 + bx + c\).
Such expressions are central to many algebraic problems.
Here are the characteristics of quadratic expressions:

• The term \(ax^2\) is the quadratic term, where \(a\) is called the leading coefficient.
• The term \(bx\) is the linear term, and \(c\) is the constant term.
• Quadratic expressions represent parabolas when graphed, and have either a U-shape (if \(a > 0\)) or an inverted U-shape (if \(a < 0\)).
• Factoring quadratic expressions is essential to identifying the roots or critical points.

The ability to factor or use the quadratic formula is crucial in manipulating and understanding these expressions. In the given inequality, the expression \(2x^2 + x - 1\) was factored to find critical points for solving the inequality.

Critical Points

Critical points in the context of inequalities and polynomials are values of \(x\) at which the expression equals zero. These points are important because they help divide the domain into intervals that can be tested.
Understanding critical points aids in analyzing the behavior of mathematical functions.
Here's why they are important:

• To find these points, set the expression equal to zero and solve for \(x\).
• For a quadratic equation, like \((2x - 1)(x + 1) = 0\), the solutions, \(-1\) and \(\frac{1}{2}\), are the critical points.
• These points help us determine where the expression switches signs.
• Testing intervals between critical points helps identify which parts satisfy the inequality.

Once identified, critical points can show where solutions start, end, or continue through infinity, making them valuable in sketching and understanding the complete graph of an expression.