Determine and show graphically the solution set of each of the following inequalities:
(a) [tex]$2-x-x^3 \geq 0$[/tex]
(b) [tex]$9-x^3\ \textgreater \ 0$[/tex]
(c) [tex]$3+x-2 x^3 \geq 0$[/tex]
(d) [tex]$2-3 x-9 x\ \textgreater \ 0$[/tex]

Solve each inequality separately by finding the roots of the corresponding equation and testing intervals.
For (a), the solution is x ≤ 1 .
For (b), the solution is x < 3 9 ​ ≈ 2.08 .
For (c), the solution is x ≤ 1.29 .
For (d), the solution is x < 6 1 ​ .

Explanation

Problem Analysis We are given four inequalities: (a) 2 − x − x 3 ≥ 0 (b) 0"> 9 − x 3 > 0 (c) 3 + x − 2 x 3 ≥ 0 (d) 0"> 2 − 3 x − 9 x > 0

The objective is to find the solution set for each inequality and represent it graphically.

Solving Inequality (a) (a) For 2 − x − x 3 ≥ 0 , let f ( x ) = 2 − x − x 3 . We need to find the roots of f ( x ) = 0 . Using a calculator tool, we find that the approximate real root is x ≈ 1.0 . To determine the intervals where f ( x ) ≥ 0 , we can test values in the intervals ( − ∞ , 1 ] and [ 1 , ∞ ) .

For x = 0 , 0"> f ( 0 ) = 2 > 0 , so the interval ( − ∞ , 1 ] is part of the solution. For x = 2 , f ( 2 ) = 2 − 2 − 2 3 = − 8 < 0 , so the interval [ 1 , ∞ ) is not part of the solution.
Therefore, the solution set is x ≤ 1 .

Solving Inequality (b) (b) For 0"> 9 − x 3 > 0 , let g ( x ) = 9 − x 3 . We need to find the roots of g ( x ) = 0 , which means x 3 = 9 . Thus, x = 3 9 ​ ≈ 2.08 . To determine the intervals where 0"> g ( x ) > 0 , we can test values in the intervals ( − ∞ , 3 9 ​ ) and ( 3 9 ​ , ∞ ) .

For x = 0 , 0"> g ( 0 ) = 9 > 0 , so the interval ( − ∞ , 3 9 ​ ) is part of the solution. For x = 3 , g ( 3 ) = 9 − 3 3 = 9 − 27 = − 18 < 0 , so the interval ( 3 9 ​ , ∞ ) is not part of the solution.
Therefore, the solution set is x < 3 9 ​ ≈ 2.08 .

Solving Inequality (c) (c) For 3 + x − 2 x 3 ≥ 0 , let h ( x ) = 3 + x − 2 x 3 . We need to find the roots of h ( x ) = 0 . Using a calculator tool, we find that the approximate real root is x ≈ 1.29 . To determine the intervals where h ( x ) ≥ 0 , we can test values in the intervals ( − ∞ , 1.29 ] and [ 1.29 , ∞ ) .

For x = 0 , 0"> h ( 0 ) = 3 > 0 , so the interval ( − ∞ , 1.29 ] is part of the solution. For x = 2 , h ( 2 ) = 3 + 2 − 2 ( 2 3 ) = 5 − 16 = − 11 < 0 , so the interval [ 1.29 , ∞ ) is not part of the solution.
Therefore, the solution set is x ≤ 1.29 .

Solving Inequality (d) (d) For 0"> 2 − 3 x − 9 x > 0 , simplify the inequality to 0"> 2 − 12 x > 0 , which gives 12 x < 2 , or x < 6 1 ​ .

Therefore, the solution set is x < 6 1 ​ .

Final Answer The solution sets for the inequalities are: (a) x ≤ 1 (b) x < 3 9 ​ ≈ 2.08 (c) x ≤ 1.29 (d) x < 6 1 ​

Examples
Understanding inequalities is crucial in many real-world scenarios. For example, in economics, companies use inequalities to determine the range of prices that maximize profit. In engineering, inequalities are used to ensure that structures can withstand certain loads or stresses. In everyday life, we use inequalities to manage budgets, ensuring that expenses do not exceed income. Inequalities help us make informed decisions and optimize outcomes in various situations.

Answered by GinnyAnswer | 2025-07-03

The inequalities yield the following solution sets: (a) x ≤ 1 , (b) x < 3 9 ​ ≈ 2.08 , (c) x ≤ 1.29 , and (d) x < 6 1 ​ .
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Answered by Anonymous | 2025-07-04