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Answer :
To solve the inequality [tex]\(2x^3 - 3x^2 - 32x > -48\)[/tex], let's follow these steps to understand it thoroughly:
1. Rearrange the inequality:
Start by moving all terms to one side of the inequality so that it becomes:
[tex]\[
2x^3 - 3x^2 - 32x + 48 > 0
\][/tex]
2. Find the critical points:
To find where the expression changes signs, we need to solve the equation:
[tex]\[
2x^3 - 3x^2 - 32x + 48 = 0
\][/tex]
The critical points, solving this cubic equation, are [tex]\(x = -4\)[/tex], [tex]\(x = \frac{3}{2}\)[/tex], and [tex]\(x = 4\)[/tex].
3. Determine the intervals:
Based on the critical points, the real number line is divided into the following intervals:
- [tex]\( (-\infty, -4) \)[/tex]
- [tex]\( (-4, \frac{3}{2}) \)[/tex]
- [tex]\( (\frac{3}{2}, 4) \)[/tex]
- [tex]\( (4, \infty) \)[/tex]
4. Test the intervals:
To determine where the inequality [tex]\(2x^3 - 3x^2 - 32x + 48 > 0\)[/tex] holds true, choose test points from each interval and substitute them back into the inequality:
- Interval [tex]\((-∞, -4)\)[/tex]: Choose test point [tex]\(x = -5\)[/tex].
- Interval [tex]\((-4, \frac{3}{2})\)[/tex]: Choose test point [tex]\(x = 0\)[/tex].
- Interval [tex]\((\frac{3}{2}, 4)\)[/tex]: Choose test point [tex]\(x = 2\)[/tex].
- Interval [tex]\((4, ∞)\)[/tex]: Choose test point [tex]\(x = 5\)[/tex].
After evaluating the expression [tex]\(2x^3 - 3x^2 - 32x + 48\)[/tex] and determining its sign in each interval, you will be able to see where it is greater than zero.
5. Conclusion:
After testing the inequalities in each interval and based on the given critical points' results, the solution to [tex]\(2x^3 - 3x^2 - 32x + 48 > 0\)[/tex] is:
- The values of [tex]\(x\)[/tex] for which the inequality holds true are within certain intervals. Hence, the solution is:
[tex]\[
x \in (-\infty, -4) \cup \left(\frac{3}{2}, 4\right)
\][/tex]
This solution describes the values for which the inequality is satisfied, taking into account the critical points where the expression changes its sign.
1. Rearrange the inequality:
Start by moving all terms to one side of the inequality so that it becomes:
[tex]\[
2x^3 - 3x^2 - 32x + 48 > 0
\][/tex]
2. Find the critical points:
To find where the expression changes signs, we need to solve the equation:
[tex]\[
2x^3 - 3x^2 - 32x + 48 = 0
\][/tex]
The critical points, solving this cubic equation, are [tex]\(x = -4\)[/tex], [tex]\(x = \frac{3}{2}\)[/tex], and [tex]\(x = 4\)[/tex].
3. Determine the intervals:
Based on the critical points, the real number line is divided into the following intervals:
- [tex]\( (-\infty, -4) \)[/tex]
- [tex]\( (-4, \frac{3}{2}) \)[/tex]
- [tex]\( (\frac{3}{2}, 4) \)[/tex]
- [tex]\( (4, \infty) \)[/tex]
4. Test the intervals:
To determine where the inequality [tex]\(2x^3 - 3x^2 - 32x + 48 > 0\)[/tex] holds true, choose test points from each interval and substitute them back into the inequality:
- Interval [tex]\((-∞, -4)\)[/tex]: Choose test point [tex]\(x = -5\)[/tex].
- Interval [tex]\((-4, \frac{3}{2})\)[/tex]: Choose test point [tex]\(x = 0\)[/tex].
- Interval [tex]\((\frac{3}{2}, 4)\)[/tex]: Choose test point [tex]\(x = 2\)[/tex].
- Interval [tex]\((4, ∞)\)[/tex]: Choose test point [tex]\(x = 5\)[/tex].
After evaluating the expression [tex]\(2x^3 - 3x^2 - 32x + 48\)[/tex] and determining its sign in each interval, you will be able to see where it is greater than zero.
5. Conclusion:
After testing the inequalities in each interval and based on the given critical points' results, the solution to [tex]\(2x^3 - 3x^2 - 32x + 48 > 0\)[/tex] is:
- The values of [tex]\(x\)[/tex] for which the inequality holds true are within certain intervals. Hence, the solution is:
[tex]\[
x \in (-\infty, -4) \cup \left(\frac{3}{2}, 4\right)
\][/tex]
This solution describes the values for which the inequality is satisfied, taking into account the critical points where the expression changes its sign.
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