91Ó°ÊÓ

The given inequality is \( \frac{x^{3}+2x^{2}}{2}<x+2 \). To eliminate the fraction, multiply every term by 2 to get rid of the denominator:\[ x^3 + 2x^2 < 2(x + 2) \].

Distribute on the right side:\[ x^3 + 2x^2 < 2x + 4 \].

Subtract \(2x\) and 4 from both sides to set the inequality to zero:\[ x^3 + 2x^2 - 2x - 4 < 0 \].

Factor the polynomial \( x^3 + 2x^2 - 2x - 4 \). Start by testing possible rational roots or factor by grouping. The polynomial can be factored as: \( (x + 2)(x^2 - 4) < 0 \).

The quadratic \( x^2 - 4 \) can be factored further:\( x^2 - 4 = (x - 2)(x + 2) \).

Now, the inequality becomes:\( (x + 2)^2 (x - 2) < 0 \).

Solve for the critical points by setting each factor equal to zero:\( x + 2 = 0 \) gives \( x = -2 \) (a repeated root).\( x - 2 = 0 \) gives \( x = 2 \).

Based on the critical points \( x = -2 \) and \( x = 2 \), determine the intervals to test: \((-\infty, -2)\), \((-2, 2)\), and \((2, \infty)\). Choose a test value from each interval and substitute it into the inequality \( (x + 2)^2 (x - 2) < 0 \) to determine where the inequality is true. For example:* Test \( x = -3 \) in \((-\infty, -2)\): \((x + 2)^2\) is positive, \((x - 2)\) is negative, result is negative (satisfies the inequality).* Test \( x = 0 \) in \((-2, 2)\): \((x + 2)^2\) is positive, \((x - 2)\) is negative, result is negative (satisfies the inequality).* Test \( x = 3 \) in \((2, \infty)\): \((x + 2)^2\) is positive, \((x - 2)\) is positive, result is positive (does not satisfy the inequality).

The inequality is satisfied in the intervals \((-\infty, -2)\) and \((-2, 2)\). These solutions can be combined as we approach \(x = -2\) without actually including it, due to the inequality being strictly less than:The final solution is \((-\infty, 2)\).