Problem 61 - Calculus Early Transcendentals

Question

$$ \begin{array}{c}{\text { Use a graph to find a number } N \text { such that }} \\ {\text { If } \quad x>N \quad \text { then } \quad\left|\frac{3 x^{2}+1}{2 x^{2}+x+1}-1.5\right|<0.05}\end{array} $$

Step-by-Step Solution

Answer

Choose \( N = 2 \). For \( x > 2 \), the function meets the condition.

Step 1: Rewrite the Expression

First, rewrite the given expression in a simpler form. We are looking for \(N\) such that \(\left|\frac{3x^2+1}{2x^2+x+1} - 1.5\right| < 0.05\). This simplifies to solving the inequality \(1.45 < \frac{3x^2+1}{2x^2+x+1} < 1.55\).

Step 2: Solve the Inequality Algebraically

Convert \(\frac{3x^2+1}{2x^2+x+1} = 1.5\) into \(0.5 \times (2x^2 + x +1) = 3x^2 + 1\), simplifying to \(x^2 -x -2 = 0\). The roots of this are \(x = 2\) and \(x = -1\). The inequality will thus include intervals outside of these roots.

Step 3: Graph the Function

Graph the function \(f(x) = \frac{3x^2+1}{2x^2+x+1}\) and the lines \(y = 1.45\) and \(y = 1.55\) to visualize where \(f(x)\) lies within the boundaries. Use a graphing tool for precision.

Step 4: Identify the Interval

Observe the graph to find where \(f(x)\) remains between 1.45 and 1.55. The function settles between these values for a set of \(x\) values, particularly where \(x > 2\).

Step 5: Confirm with Graph

Verify with the graph that starting at \(x > 2\), the function \(f(x)\) stays within the values of 1.45 and 1.55.

Key Concepts

Understanding Inequalities

Inequalities are mathematical expressions involving the symbols \(<, >, \leq, \geq\), which help us compare two values or expressions. When solving inequalities, we seek ranges or sets of values that satisfy the condition. A critical aspect in dealing with inequalities is maintaining the truth of the inequality across transformations, such as addition or multiplication. When multiplying or dividing by a negative number, remember to flip the inequality sign.

In the exercise at hand, we're dealing with a compound inequality: \(1.45 < \frac{3x^2+1}{2x^2+x+1} < 1.55\). This form tells us that we're looking for the range of \(x\) values where the expression fits between two bounds. To solve it, it often helps to break it into two simple inequalities:

• \(\frac{3x^2+1}{2x^2+x+1} > 1.45\)
• \(\frac{3x^2+1}{2x^2+x+1} < 1.55\)

After finding the solution to each inequality, these help us determine the specific interval of \(x\) that satisfies the entire condition.

The Role of Quadratic Equations

Quadratic equations are expressions in the form \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are constants. They play a pivotal role in the context of the exercise, as they arise when equating our given function \(\frac{3x^2+1}{2x^2+x+1}\) to a constant. When simplified, it transforms into a quadratic equation like \(x^2 - x - 2 = 0\).

Solving quadratic equations can be done using several methods, such as factoring, completing the square, or using the quadratic formula. For this exercise, factoring gives the solutions \(x = 2\) and \(x = -1\). These solutions are critical in defining intervals for our inequality, signaling transitions where our function changes direction or behavior. By understanding the solutions of the quadratic equation, one can ascertain the intervals to test in an inequality context.

Graphing Functions for Insight

Graphing functions is a pivotal technique in visualizing mathematical concepts, especially when analyzing limits and inequalities. By graphing, we translate abstract algebraic expressions into visual representations that can help spot patterns, behaviors, and solutions easily. For our specific task, graphing the function \(f(x) = \frac{3x^2+1}{2x^2+x+1}\) alongside the boundaries \(y = 1.45\) and \(y = 1.55\) brings clarity.

On a graph, the intersection points where \(f(x)\) crosses \(1.45\) and \(1.55\) are significant. They indicate where the function exits or enters the desired range between these two bounds. In our solution, identifying the interval \(x > 2\) is made simpler by observing the graph where \(f(x)\) consistently lies between the two horizontal lines representing the limits. This visual insight complements the algebraic approach, making it easier to understand why \(x\) must be greater than 2 for the initial inequality condition.