Problem 79 - A Graphical Approach to College Algebra

Question

Solve each equation graphically. $$|x+1|+|x-6|=11$$

Step-by-Step Solution

Answer

Solutions are \(x = -3\) and \(x = 8\).

Step 1: Understand the Absolute Value Equation

The given equation is \(|x+1| + |x-6| = 11\). This involves two absolute value expressions. We need to find the values of \(x\) that satisfy this equation. To solve this graphically, consider each piecewise linear function formed by these absolute values.

Step 2: Identify Critical Points

The critical points for the absolute value expressions are \(x = -1\) and \(x = 6\). At these points, the expressions inside the absolute values change sign. Hence, we will split the real number line into three intervals: \((-\infty, -1)\), \([-1, 6)\), and \([6, \infty)\).

Step 3: Analyze and Graph Each Interval

For \((-\infty, -1)\), both expressions are negative:- \(|x+1| = -(x+1)\)- \(|x-6| = -(x-6)\).The equation becomes: \(-(x+1) - (x-6) = 11\) i.e., \(-2x+5 = 11\). Solve for \(x\) to find \(x = -3\).For \([-1, 6)\), \(|x+1| = x+1\) and \(|x-6| = -(x-6)\).The equation becomes:\((x+1) - (x-6) = 11\) i.e., \(7 = 11\), which is not possible.For \([6, \infty)\), both expressions are positive:- \(|x+1| = x+1\)- \(|x-6| = x-6\).The equation becomes:\((x+1) + (x-6) = 11\) i.e., \(2x - 5 = 11\). Solve for \(x\) to find \(x = 8\).

Step 4: Plot and Verify the Solutions

Graph the simplified equations for each interval:- A line through \(x = -3\) for \((-\infty, -1)\). This is a potential solution since it satisfies the equation.- No solution in \([-1, 6)\) since \(7 eq 11\).- A line through \(x = 8\) for \([6, \infty)\). This too is a solution as it satisfies the equation.Thus, graphically verify these intersections at \(x = -3\) and \(x = 8\).

Step 5: Conclusion

The pieces of the function for each interval show that the only solutions are where the piecewise lines intersect to meet the condition \(|x+1| + |x-6| = 11\). Therefore, the solutions are \(x = -3\) and \(x = 8\).

Key Concepts

Graphical Solution

Solving absolute value equations graphically involves transforming the equation into visually understandable graphical lines or curves. The equation \(|x+1| + |x-6| = 11\) can be represented on a graph by plotting different lines for the absolute value parts within specified intervals. This is because absolute values form V-shaped graphs. By identifying where these plotted lines intersect with each other or with a constant line, we can determine the solutions to the equation.

In this instance, by transforming and plotting the lines for each interval, we observe two points where the curves meet the horizontal line at 11, specifically at \(x = -3\) and \(x = 8\). Thus, these intersections reveal the solutions to the equation. This visual method confirms numeric solutions obtained through algebraic calculations and provides a straightforward way to comprehend where the solution lies on a number line.

Graphical solutions help in understanding how values vary, especially when dealing with piecewise characteristics like those in absolute value expressions.

Piecewise Functions

A piecewise function is one that is defined by different expressions based on different intervals. In the case of an absolute value equation like \(|x+1| + |x-6| = 11\), we analyze it as a collection of linear segments. Each segment corresponds to a different combination of the inner expressions being positive or negative.

This equation is broken into pieces around the critical points, \(x = -1\) and \(x = 6\), which define the intervals:

• For \((-\infty , -1)\), both expressions are negative, forming an interval-specific function: \(-(x+1) - (x-6)\).
• For \([-1, 6)\), one expression is positive, the other is negative, hence the function becomes \((x+1) - (x-6)\).
• For \([6, \infty )\), both are positive, and the function is \((x+1) + (x-6)\).

By integrating the solutions from each of these distinct segments, we conform the behavior of the piecewise function across the entire number line, capturing all possible values of \(x\) that satisfy the original equation.

Critical Points

Critical points in the context of absolute value equations are the values where the expression within the absolute value equals zero. They are important because they determine the intervals over which the behavior of the function changes.

For the equation \(|x+1| + |x-6| = 11\), the critical points are \(x = -1\) and \(x = 6\). At these points:

• \(x+1 = 0\) implies \(x = -1\),
• \(x-6 = 0\) implies \(x = 6\).

These points split the number line into sections where the sign (positive or negative) of each part changes. As a result, the method of calculating the function’s absolute value changes as well. They allow us to create piecewise segments that reflect the real numerical behavior within each section. By checking for solutions within these sections, we can accurately determine where the absolute value equation holds true.

Thus, identifying critical points is a key step in solving absolute value equations, ensuring that all sections where potential solutions may lie are properly analyzed.