Solving The Inequality -v^3 ≤ V^2 - 132v In Interval Notation

This article provides a detailed, step-by-step guide on how to solve the inequality $-v^3 \leq v^2 - 132v$. Inequalities like this are fundamental in algebra and calculus, and mastering their solution is crucial for understanding more advanced mathematical concepts. We will explore the necessary steps, including rearranging the inequality, factoring, finding critical points, and using interval testing to determine the solution set. The final answer will be expressed in interval notation.

Understanding the Inequality

At its core, solving an inequality involves finding the range of values for a variable that makes the inequality statement true. In this case, we are looking for the values of v that satisfy the inequality $-v^3 \leq v^2 - 132v$. This is a polynomial inequality, specifically a cubic inequality, which means the highest power of the variable v is 3. Polynomial inequalities can often be solved by bringing all terms to one side, factoring, and then analyzing the intervals created by the roots. This systematic approach ensures we capture all possible solutions.

Rearranging the Inequality

The first step in solving this inequality is to rearrange it so that all terms are on one side, making the other side zero. This allows us to compare the polynomial expression to zero and identify the intervals where the inequality holds. We start with:

$-v^3 \leq v^2 - 132v$

To move all terms to the left side, we add $-v^2$ and $+132v$ to both sides of the inequality:

$-v^3 - v^2 + 132v \leq 0$

This rearrangement is crucial because it sets up the inequality for factoring, which is a key technique for finding the critical points. By having zero on one side, we can easily analyze the sign of the polynomial expression on different intervals.

Factoring the Polynomial

Now that we have the inequality in the form $-v^3 - v^2 + 132v \leq 0$, the next step is to factor the polynomial expression. Factoring helps us identify the roots, which are the points where the expression equals zero. These roots will be the critical points that divide the number line into intervals.

First, we can factor out a common factor of $-v$ from the expression:

$-v(v^2 + v - 132) \leq 0$

Next, we need to factor the quadratic expression $v^2 + v - 132$. We are looking for two numbers that multiply to -132 and add to 1. These numbers are 12 and -11. Thus, we can factor the quadratic as:

$v^2 + v - 132 = (v + 12)(v - 11)$

So, the fully factored inequality is:

$-v(v + 12)(v - 11) \leq 0$

Factoring the polynomial is a critical step. It transforms a complex inequality into a product of simpler factors, each of which can be easily analyzed. The roots of these factors are the critical points that determine the intervals we need to test.

Identifying Critical Points

The critical points are the values of v that make the expression $-v(v + 12)(v - 11)$ equal to zero. These points divide the number line into intervals, and the sign of the expression within each interval will be consistent. To find the critical points, we set each factor equal to zero:

$-v = 0 \implies v = 0$

$v + 12 = 0 \implies v = -12$

$v - 11 = 0 \implies v = 11$

Therefore, the critical points are $v = -12$, $v = 0$, and $v = 11$. These points are the boundaries of the intervals we will test to find the solution set of the inequality. Critical points are essential because they are the only points where the expression can change its sign. By identifying these points, we can efficiently analyze the intervals.

Interval Testing

With the critical points identified, the next step is to test intervals on the number line to determine where the inequality $-v(v + 12)(v - 11) \leq 0$ holds true. The critical points $-12$, $0$, and $11$ divide the number line into four intervals: $(-\infty, -12)$, $(-12, 0)$, $(0, 11)$, and $(11, \infty)$. We will choose a test value within each interval and evaluate the expression to see if it is less than or equal to zero.

Test Intervals

1. Interval $(-\infty, -12)$: Choose $v = -13$

   $-(-13)(-13 + 12)(-13 - 11) = 13(-1)(-24) = 312$. Since $312 > 0$, this interval does not satisfy the inequality.

2. Interval $(-12, 0)$: Choose $v = -1$

   $-(-1)(-1 + 12)(-1 - 11) = 1(11)(-12) = -132$. Since $-132 \leq 0$, this interval satisfies the inequality.

3. Interval $(0, 11)$: Choose $v = 1$

   $-(1)(1 + 12)(1 - 11) = -1(13)(-10) = 130$. Since $130 > 0$, this interval does not satisfy the inequality.

4. Interval $(11, \infty)$: Choose $v = 12$

   $-(12)(12 + 12)(12 - 11) = -12(24)(1) = -288$. Since $-288 \leq 0$, this interval satisfies the inequality.

Including Critical Points

Since the inequality is $-v(v + 12)(v - 11) \leq 0$ (less than or equal to zero), we include the critical points in the solution set. The critical points are $v = -12$, $v = 0$, and $v = 11$.

Expressing the Solution in Interval Notation

Now that we have tested the intervals and included the critical points, we can express the solution set in interval notation. The intervals that satisfy the inequality are $(-12, 0)$ and $(11, \infty)$. Including the critical points, we have the intervals $[-12, 0]$ and $[11, \infty]$.

Final Solution

The solution to the inequality $-v^3 \leq v^2 - 132v$ in interval notation is:

$[-12, 0] \cup [11, \infty]$

This notation indicates that the solution set includes all values of v between -12 and 0 (inclusive), as well as all values of v greater than or equal to 11. The union symbol $\cup$ combines these two intervals into a single solution set.

Conclusion

Solving inequalities such as $-v^3 \leq v^2 - 132v$ requires a systematic approach, including rearranging the inequality, factoring the polynomial expression, identifying critical points, testing intervals, and expressing the solution in interval notation. By following these steps, we can accurately determine the range of values that satisfy the inequality. Understanding and mastering these techniques is crucial for success in algebra and calculus. Remember to always double-check your work and ensure that your solution makes sense in the context of the original problem. This comprehensive guide provides a clear pathway to solving similar inequalities, enhancing your mathematical skills and problem-solving abilities.

Key takeaways include:

• Rearranging the inequality to have zero on one side.
• Factoring the polynomial expression to find critical points.
• Identifying critical points as the roots of the factored expression.
• Testing intervals to determine where the inequality holds.
• Including critical points in the solution set if the inequality includes "equal to."
• Expressing the solution in interval notation.

By consistently applying these steps, you can confidently solve a wide range of polynomial inequalities.