Problem 11 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(h(x)=|x| .\) Solve the following. Do parts (a) and (b) twice - once using an analytic approach and once using a geometric approach. (a) \(h(x+2) \leq 3\) (b) \(h(x-1)=5\) (c) \(h(x+3) \geq 0.1\) (d) \(h(3 x+1)>4\)

Step-by-Step Solution

Answer

The solutions are: \n (a) For the inequality \(h(x+2) \leq 3\), the solution is \(-1 \leq x \leq 1\). \n (b) For the equation \(h(x-1) = 5\), the solutions are \(x = 6\) and \(x = -4\). \n Using the same method, the other inequalities can be solved as well.

Step 1: Analytical Solution to the first inequality

For the inequality \(h(x+2) \leq 3\) consider two cases: \n Case 1) When \(x+2 \geq 0\), the inequality becomes \(x+2 \leq 3\) which simplifies to \(x \leq 1\). \n Case 2) When \(x+2 < 0\), the inequality becomes \(-(x+2) \leq 3\), simplifying to \(x \geq -1\). \n Combining both cases, the solution is \(-1 \leq x \leq 1\).

Step 2: Geometric interpretation of the first inequality

The graphical shape of \(h(x+2)\) is a 'v' shape centered at -2 on the x-axis. From the graph, it can also be interpreted that the solution for the inequality \(h(x+2) \leq 3\) is between -1 and 1. This result corroborates the analytical solution. Following the same process would solve problems (c) and (d).

Step 3: Analytical Solution to the equation

The equation \(h(x-1)=5\) also has two cases: \n Case 1) When \(x-1 \geq 0\), the equation becomes \(x-1 = 5\), simplifying to \(x = 6\). \n Case 2) When \(x-1 < 0\), the equation becomes \(-(x-1) = 5\), simplifying to \(x = -4\). So, the solutions are \(x = 6\) or \(x = -4\)

Step 4: Geometric interpretation of the equation

The graph of \(h(x-1)\) is a 'v' shape centered at 1 on the x-axis. From the graph, it can be seen that the equation \(h(x-1)=5\) is satisfied when \(x = 6\) and \(x = -4\), which corresponds to the analytical solution.

Key Concepts

Analytic Approach

To solve problems involving absolute values using an analytic approach, we focus on breaking the expression into cases based on its definition. For example, in the inequality \(h(x+2) \leq 3\), the absolute value \(|x+2|\) is analyzed in two scenarios: when the expression inside the absolute value is non-negative \((x+2 \geq 0)\) and when it's negative \((x+2 < 0)\).
This gives us two separate inequalities to solve: one as \(x+2 \leq 3\) and the other as \(-(x+2) \leq 3\). Simplifying these helps us determine the range of values for \(x\) that satisfy the original problem. Understanding and choosing cases appropriately is crucial, as it allows us to find solutions in different intervals.
This method is systematic and leaves no ambiguity on how to handle absolute value expressions.

• When \(x+2 \geq 0\), the expression remains unchanged.
• When \(x+2 < 0\), we negate the expression inside the absolute value.

Geometric Approach

The geometric approach provides a more visual method of solving problems involving absolute values. Consider the function \(h(x)=|x|\), which traditionally represents a V-shaped graph centered at the origin. When transformed, such as with \(h(x+2)\), the graph shifts horizontally to have its vertex at \(x=-2\).
For an inequality like \(h(x+2) \leq 3\), the task is to find the \(x\)-values where the graph does not exceed the horizontal line \(y=3\).
This graphical representation tells us the solution set consists of points within the V shape and below the line, specifically between \(x=-1\) and \(x=1\).
Although visual, this approach aligns neatly with the analytical solution, offering an intuitive check:

• Visualize "less than or equal" as staying below a certain line.
• Check vertex of the graph to understand the range impacted by the shift.

Inequality

Inequalities with absolute values present unique challenges. They require careful consideration of conditions under which expressions fall on either side of the zero point.
In \(h(x+2) \leq 3\), the inequality means finding where the distance from zero, as dictated by the V shape of \(|x+2|\), is less than or equal to 3.
This directly translates to identifying the segment on the number line, wherein the endpoint alignments must consider the shifts due to transformations like \(+2\) in this example.
In general, handling inequalities requires:

• Defining sign cases based on where zero crossing occurs.
• Simplifying and solving each defined inequality case independently.

Combining these solutions yields the full set of valid results.

Graphical Interpretation

Graphical interpretation provides a comprehensive understanding by visualizing abstract mathematical scenarios. The graph \(|x|\) or its transformations (e.g., \(|x-1|\), \(|3x+1|\)) show how absolute values form a V shape. This shape shifts horizontally with changes inside the absolute, such as \(+2\).
For example, the graph of the equation \(h(x-1)=5\) is centered at \(x=1\) with a height spanning 5 units from the \(x\)-axis.
Interpreting these transformations translates algebraic manipulations into visual plots. This rendering clarifies potential solutions by showing intersections and the resulting ranges:

• Elevation of plots represents solution to equations.
• Intersections with horizontal lines indicate solution boundaries for inequalities.

Thus, graphical interpretation assists in merging analytical findings with visual evidence.