Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Express the solution set of the given inequality in interval notation and sketch its graph. $$ x^{2}+2 x-12<0 $$

Step-by-Step Solution

Answer

The solution set is \((-4, 3)\).

Step 1: Factor the Quadratic Expression

To solve the inequality \(x^2 + 2x - 12 < 0\), we need to factor the quadratic expression. We look for two numbers that multiply to \(-12\) and add to \(2\).The factors are \((x+4)(x-3)\) since \(4 \times -3 = -12\) and \(4 + (-3) = 1\). Thus, the inequality becomes \((x+4)(x-3) < 0\).

Step 2: Find the Critical Points

The critical points occur where the expression equals zero. Set \((x+4)(x-3) = 0\) and solve for \(x\):- \(x+4=0\) gives \(x=-4\)- \(x-3=0\) gives \(x=3\)Critical points are \(-4\) and \(3\).

Step 3: Determine the Test Intervals

Use the critical points to divide the number line into intervals. The intervals are :1. \((-\infty, -4)\)2. \((-4, 3)\)3. \((3, \infty)\)

Step 4: Test Each Interval

Select a test point from each interval and plug it into the inequality \((x+4)(x-3) < 0\):1. For \((-\infty, -4)\), choose \(-5\): \((-5+4)(-5-3) = (-1)(-8) = 8\), not less than 0.2. For \((-4, 3)\), choose \(0\): \((0+4)(0-3) = (4)(-3) = -12\), less than 0.3. For \((3, \infty)\), choose \(4\): \((4+4)(4-3) = (8)(1) = 8\), not less than 0.

Step 5: Write the Solution in Interval Notation

Since the expression is less than zero only over the interval \((-4, 3)\), the solution set is \((-4, 3)\).

Step 6: Sketch the Graph

On a number line, draw open circles at \(-4\) and \(3\) to indicate these points are not included, and shade the region between \(-4\) and \(3\) to represent \((-4, 3)\), where the inequality is satisfied.

Key Concepts

Quadratic Expression

A quadratic expression is a polynomial of degree two, generally expressed as \( ax^2 + bx + c \). When solving inequalities with quadratic expressions, a common technique is to factor the quadratic to make analysis simpler. This involves finding two numbers that multiply to the constant term (here, \(-12\)) and add to the linear coefficient (here, \(2\)). Once factored, the quadratic expression \( x^2 + 2x - 12 \) becomes \( (x+4)(x-3) \). This new form is often easier to work with when solving inequalities because it breaks down the problem into more manageable parts. Each term in the factored form represents a potential critical point where the inequality changes its behavior.

Critical Points

In analyzing quadratic inequalities, critical points play a vital role. These are the points where the expression equals zero. For \( (x+4)(x-3) = 0 \), solving the equation gives the critical points \( x = -4 \) and \( x = 3 \).

• The critical points divide the number line into distinct segments or intervals.
• These intervals help determine where the expression is greater than, less than, or equal to zero.

Critical points are essential as they give us the boundaries of these segments, allowing us to test and analyze different intervals of values.

Interval Notation

Interval notation is a way of describing a set of numbers, typically representing the solution set of an inequality. Here, the solution obtained after analyzing the intervals is expressed using interval notation, \((-4, 3)\). This particular format communicates:

• Open circles at points \(-4\) and \(3\) indicate these endpoints are not included in the solution.
• The parentheses \(( )\) further confirm that the interval does not include \(-4\) and \(3\).

Interval notation provides a visual and efficient way to present solutions, eliminating the need for lengthy explanations while clearly conveying the range of values where the inequality holds true.

Test Intervals

Testing intervals is a critical step after identifying the critical points. Each interval between and beyond these points needs to be tested to determine where the inequality holds:- For the interval \((-\infty, -4)\), a test point like \(-5\) yields a positive product, not satisfying the \(< 0\) condition.- The interval \((-4, 3)\) with a test point like \(0\) results in a negative product, meeting the inequality condition.- Finally, in the interval \((3, \infty)\), using a test point like \(4\) gives a positive result again, not satisfying the inequality.This process ensures that only the interval \((-4, 3)\) is included in the final solution, as it is the only interval that satisfies the quadratic inequality \((x+4)(x-3) < 0\). Choosing appropriate test points and properly evaluating them is crucial for correctly solving such inequalities.