Problem 23 - The Calculus with Analytic Geometry

Question

In Exercises 15 through 26 , find the solution set of the given inequality, and illustrate the solution on the real number line. $$ |9-2 x| \geq|4 x| $$

Step-by-Step Solution

Answer

\(\left[-\frac{9}{2}, \frac{3}{2}\right]\)

Step 1: Square Both Sides

Since both sides are non-negative, \(|9-2x| \geq |4x|\) iff \((9-2x)^2 \geq (4x)^2\):
\(81 - 36x + 4x^2 \geq 16x^2\)
\(81 - 36x - 12x^2 \geq 0\)
\(12x^2 + 36x - 81 \leq 0\)

Step 2: Solve the Quadratic Inequality

\(12x^2 + 36x - 81 = 0\)
\(x = \frac{-36 \pm \sqrt{1296 + 3888}}{24} = \frac{-36 \pm \sqrt{5184}}{24} = \frac{-36 \pm 72}{24}\)
\(x = \frac{36}{24} = \frac{3}{2}\) or \(x = \frac{-108}{24} = -\frac{9}{2}\)

Step 3: State the Solution

Since the parabola opens upward, the inequality \(12x^2 + 36x - 81 \leq 0\) holds for \(-\frac{9}{2} \leq x \leq \frac{3}{2}\).
Solution set: \(\left[-\frac{9}{2}, \frac{3}{2}\right]\)

Key Concepts

Absolute Value Inequalities

Absolute value inequalities involve expressions within absolute value signs. The absolute value \(|x|\) of a number is its distance from zero on the number line, regardless of direction. So, the expression \(|9-2x|\) represents how far \((9-2x)\) is from zero.
In solving \(|9-2x| \geq |4x|\), we deal with two primary cases:

• Positive case: \(9-2x \geq 4x\)
• Negative case: \(-(9-2x) \geq 4x\)

Solving these separately will give us the intervals where our inequality holds true.

Solution Set

The solution set of an inequality includes all the values that satisfy the condition. Breaking down \(|9-2x| \geq |4x|\)\r:

• For \(9-2x \geq 4x\), isolating \(x\) gives \(9 \geq 6x \Rightarrow x \leq \frac{3}{2}\).
• For \(-(9-2x) \geq 4x\) simplifies to \(2x-9 \geq 4x \Rightarrow -9 \geq 2x \Rightarrow x \leq -9\).

So, the combined solution set is \(x \leq -9\) or \(x \leq \frac{3}{2}\).
Since -9 is already within the range \(x \leq \frac{3}{2}\), the final solution is \(x \leq \frac{3}{2}\).

Real Number Line

To visualize the solution, we use a real number line. The number line helps us graphically represent the inequality and the solution set.
Here’s how we do it:

• Mark the critical point \(\frac{3}{2}\), which is 1.5.
• Since \(x \leq \frac{3}{2}\), shade the line to the left of \(\frac{3}{2}\).
• This shaded part represents all the numbers less than or equal to \(\frac{3}{2}\).

By drawing the solution on the number line, one can easily see the range of values that solve the inequality, making it visually clear.