Problem 60 - A Graphical Approach to College Algebra

Question

Graph the solution set of each system of inequalities by hand. $$\begin{array}{r} \ln x-y \geq 1 \\ x^{2}-2 x-y \leq 1 \end{array}$$

Step-by-Step Solution

Answer

Graph \( y = \ln x - 1 \), shade below; graph \( y = x^2 - 2x - 1 \), shade above; identify overlapping region.

Step 1: Rewrite the Inequalities for Graphing

For the inequality \( \ln x - y \geq 1 \), rewrite it as \( y \leq \ln x - 1 \). This represents all points below or on the line \( y = \ln x - 1 \). For the second inequality \( x^2 - 2x - y \leq 1 \), rewrite it as \( y \geq x^2 - 2x - 1 \). This represents all points above or on the parabola \( y = x^2 - 2x - 1 \).

Step 2: Graph the Boundary of the Logarithmic Inequality

Graph the logarithmic function \( y = \ln x - 1 \). Since \( y \leq \ln x - 1 \), shade the region below this curve. Remember that the natural logarithm \( \ln x \) is only defined for \( x > 0 \), hence the graph and shading start just to the right of the y-axis.

Step 3: Graph the Parabolic Boundary

Graph the parabola defined by \( y = x^2 - 2x - 1 \). Since \( y \geq x^2 - 2x - 1 \), shade the region above this parabola. Focus on achieving an accurate sketch to visualize where this parabola sits on the graph with respect to the logarithmic function.

Step 4: Identify the Solution Region

Determine the overlapping region from the shaded areas in Step 2 and Step 3. The solution set is the region where both inequalities overlap; it is where the points satisfy both conditions simultaneously.

Step 5: Verify the Shaded Region

Pick a test point within the overlap region to ensure it satisfies both inequalities. For example, check that a point like \((x, y)\) within the overlapping region satisfies both \( y \leq \ln x - 1 \) and \( y \geq x^2 - 2x - 1 \).

Key Concepts

Logarithmic Functions

A logarithmic function, often written as \( y = \ln x \), is the inverse of an exponential function. The natural logarithm \( \ln x \) gives you the power to which the base "e" must be raised to get "x". Here, the key trait to remember is that it is defined only for \( x > 0 \). This means when graphing a logarithmic inequality, like \( y \leq \ln x - 1 \), you start plotting right next to the y-axis and move to the right.

• The graph of \( y = \ln x - 1 \) shifts the standard \( \ln x \) curve down by 1 unit.
• All points below this curve are solutions to \( y \leq \ln x - 1 \). This area needs to be shaded on the graph.

Understanding these characteristics helps you not only graph but also pinpoint the behavior and limitations of logarithmic functions in inequalities.

Parabolic Equations

Parabolic equations are often encountered in the form \( y = ax^2 + bx + c \), which plot to form a U-shaped curve called a parabola. For our problem, we examine the parabola given by \( y = x^2 - 2x - 1 \).

• The vertex, or the highest or lowest point, of this parabola is critical for graphing. You can find it by using \( x = -\frac{b}{2a} \).
• In this inequality, \( y \geq x^2 - 2x - 1 \), points above or on the parabola are part of the solution.

The negative sign of the linear term \(-2x\) causes a horizontal shift of the parabola. By accurately sketching this parabola, we can see how it interacts with the logarithmic curve, marking where shading should occur.

Solution Set of Inequalities

To solve a system of inequalities, graph each function to find where their solution sets overlap. This overlap, or intersection, is where both inequalities are true simultaneously. - First, graph the region below \( y = \ln x - 1 \).- Then, graph the region above the parabola \( y = x^2 - 2x - 1 \).The critical area to focus on is the intersection of these shaded regions. It forms the final solution set that satisfies both inequalities. Verifying this with a test point from the overlap ensures no mistakes in identifying the solution area. This graphical approach to solving systems of inequalities provides a visual understanding of how different mathematical expressions relate in real space.