Problem 22 - Precalculus: Mathematical for Calculus

Question

Show that \(100 n \leq n^{2}\) for all \(n \geq 100\)

Step-by-Step Solution

Answer

The inequality 100n ≤ n^2 holds for all n ≥ 100 as n(n - 100) is non-negative.

Step 1: Understand the Inequality

The problem asks us to demonstrate that the inequality \(100n \leq n^{2}\) holds true for all \(n\) greater than or equal to 100. This means that starting at \(n = 100\), and for every value greater than 100, the expression \(n^2 - 100n\) should be non-negative.

Step 2: Rewrite the Inequality

Rewrite the inequality in terms of a single quadratic expression: \(n^2 - 100n \geq 0\). This is now a standard inequality problem where we need to find the range of \(n\) where it holds true.

Step 3: Solve the Quadratic Equation

First, solve the equation \(n^2 - 100n = 0\). Factoring gives us \(n(n - 100) = 0\). This produces the solutions \(n = 0\) and \(n = 100\). These are the critical points of the quadratic expression.

Step 4: Test the Intervals

Test the intervals determined by the critical points (\(-\infty, 0\), \(0, 100\), and \(100, \infty\)) to see where the inequality is satisfied. Specifically, check the interval \([100, \infty)\) since we are interested in \(n \geq 100\). Choose \(n = 101\) as a test point: \(101^2 - 100 \cdot 101 = 101 - 100 = 1\). Since 1 is positive, the inequality holds for \(n > 100\).

Step 5: Conclude

Since \(n(n-100)\) is positive for \(n > 100\), and both critical points (\(n=0\) and \(n=100\)) solve the original equation \(n^2 - 100n = 0\), we conclude that \(100n \leq n^2\) for all \(n \geq 100\). This confirms the inequality holds in this interval.

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These are called quadratic because the highest degree of exponent is two, which means the graph of this equation will be a parabola. In our exercise, we deal with the quadratic inequality \(n^2 - 100n \geq 0\). Here, \(a = 1\), \(b = -100\), and \(c = 0\). The aim is to show that this inequality holds true for all \(n\) greater than or equal to 100, which translates to solving the inequality rather than simply solving for \(n\) in \(n^2 - 100n = 0\). Quadratic equations require special methods to find solutions, either by factoring, using the quadratic formula, or completing the square. But first, it is essential to understand the graph behavior, which can provide insight into the nature of solutions, especially for inequalities.

Factoring

Factoring is a method used to solve quadratic equations. It involves breaking down the equation into a product of simpler expressions, called factors, that when multiplied together give the original expression. In our exercise, we start by trying to solve the equation \(n^2 - 100n = 0\) by factoring.The equation can be factored as \(n(n - 100) = 0\).

• This tells us that either \(n = 0\) or \(n = 100\) makes the equation true.
• These values are the roots or solutions of the equation and are known as critical points.

By setting each factor equal to zero, you find the values of \(n\) that make \(n^2 - 100n = 0\). These are important because they help in dividing the number line into intervals for testing the inequality, which is crucial for proving \(n^2 - 100n \geq 0\) for \(n \geq 100\). Factoring is not only useful in finding roots but also vital in interval testing, giving insight into how quadratic expressions behave over different ranges.

Critical Points

Critical points in the context of inequalities for quadratic equations refer to the values of \(n\) where the quadratic expression equals zero. These are important to identify because they divide the number line into intervals that can be tested to see where the inequality holds true.For \(n^2 - 100n = 0\), we found the critical points to be \(n = 0\) and \(n = 100\).

• These points were derived through factoring the equation into \(n(n - 100)\).
• The critical points represent where the parabola intersects the \(n\)-axis.

Because our aim is to show \(100n \leq n^2\) for all \(n \geq 100\), the critical point at \(n = 100\) is particularly important. It transitions the quadratic expression from negative or zero to positive, indicating where the inequality starts holding true. By identifying these points, we can perform interval testing to affirm our inequality statement in the desired range.

Interval Testing

Interval testing is a method used to determine where a given inequality holds true. After establishing the critical points of a quadratic equation, the next step is to test values in each interval divided by these points.In our case, the critical values of \(n\) (from factoring \(n^2 - 100n = 0\)) give intervals:

• \((-\infty, 0)\),
• \((0, 100)\), and
• \([100, \infty)\)

For the exercise, we are interested only in the interval \([100, \infty)\) due to the inequality constraint \(n \geq 100\). By selecting \(n = 101\) as a test point:

• \((101)^2 - 100 \times 101 = 101\).
• This result is positive, confirming the inequality is satisfied in this interval.

This confirms that the quadratic expression remains non-negative for all \(n \geq 100\). Testing each interval helps verify whether the inequality holds true across the desired range, providing a clear picture of its behavior over different sections of the number line.