Key Concepts and Formulas

• Monotonicity and the First Derivative: A function $f(x)$ is increasing on an interval if $f'(x) \ge 0$ for all $x$ in that interval.
• Piecewise Functions: We need to consider the derivative of each piece of the function separately.
• Continuity: While not explicitly checked here, continuity is important to ensure that the function is well-behaved at the transition points. In this problem, the derivative's sign will drive the increasing/decreasing nature, so the endpoint inclusion (or exclusion) is governed by whether the derivative is positive or zero.

Step-by-Step Solution

Step 1: Find the derivative of each piece of the function

We differentiate each part of the piecewise function $f(x)$: f(x) = \left\{ {\matrix{ { - 55x,} & {if\,x < - 5} \cr {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr } } \right.

• For $x < -5$, $f'(x) = -55$.
• For $-5 \le x \le 4$, $f'(x) = 6x^2 - 6x - 120$.
• For $x > 4$, $f'(x) = 6x^2 - 6x - 36$.

Thus, f'(x) = \left\{ {\matrix{ { - 55,} & {if\,x < - 5} \cr {6{x^2} - 6x - 120,} & {if\, - 5 < x < 4} \cr {6{x^2} - 6x - 36,} & {if\,x > 4,} \cr } } \right.

Step 2: Determine where each piece of the derivative is greater than or equal to zero

We want to find the intervals where $f'(x) \ge 0$.

• For $x < -5$, $f'(x) = -55 < 0$. So, $f(x)$ is decreasing on $(-\infty, -5)$.

• For $-5 < x < 4$, $f'(x) = 6x^2 - 6x - 120 = 6(x^2 - x - 20) = 6(x-5)(x+4)$. We need to find where $6(x-5)(x+4) \ge 0$. This inequality holds when $x \le -4$ or $x \ge 5$. However, we are only considering the interval $-5 < x < 4$. So, in this interval, $f'(x) \ge 0$ when $-5 < x \le -4$. Thus, $f(x)$ is increasing on $(-5, -4)$.

• For $x > 4$, $f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2)$. We need to find where $6(x-3)(x+2) \ge 0$. This inequality holds when $x \le -2$ or $x \ge 3$. However, we are only considering the interval $x > 4$. So, in this interval, $f'(x) > 0$ when $x > 4$. Thus, $f(x)$ is increasing on $(4, \infty)$.

Step 3: Combine the intervals where f(x) is increasing

From the previous steps, $f(x)$ is increasing on $(-5, -4)$ and $(4, \infty)$. At $x = -5$, $f'(x) = -55 < 0$. At $x = 4$, $f'(x) = 6(4^2) - 6(4) - 36 = 6(16) - 24 - 36 = 96 - 60 = 36 > 0$.

Therefore, $f(x)$ is increasing on $(-5, -4) \cup (4, \infty)$. However, the given answer is $(-5, \infty)$. Let's re-examine the interval $-5 \le x \le 4$. We have $f'(x) = 6(x-5)(x+4)$. Since we are considering the interval $-5 < x < 4$, $x-5 < 0$. For $-5 < x < -4$, we have $x+4 < 0$, so $f'(x) = 6(x-5)(x+4) > 0$, which means that $f$ is increasing on $(-5, -4)$. For $-4 < x < 4$, we have $x+4 > 0$, so $f'(x) = 6(x-5)(x+4) < 0$, which means that $f$ is decreasing on $(-4, 4)$. When $x = -5$, $f'(x) = -55 < 0$. When $x = 4$, $f'(x) = 6(16) - 6(4) - 120 = 96 - 24 - 120 = -48 < 0$. When $x > 4$, $f'(x) = 6(x-3)(x+2)$, so $f'(x) > 0$ for $x > 4$. Then the function is increasing on $(-5, -4) \cup (4, \infty)$.

Let's re-examine the derivative for $x < -5$. $f'(x) = -55 < 0$. Thus, $f(x)$ is decreasing for $x < -5$. For $-5 \le x \le 4$, we have $f'(x) = 6x^2 - 6x - 120$. At $x = -5$, $f'(-5) = 6(25) - 6(-5) - 120 = 150 + 30 - 120 = 60 > 0$. So the function should be increasing at $x = -5$. $f'(x) = 6(x-5)(x+4)$. So $f'(x) \ge 0$ when $x \le -4$ or $x \ge 5$. For $x > 4$, $f'(x) = 6x^2 - 6x - 36$. At $x = 4$, $f'(4) = 6(16) - 6(4) - 36 = 96 - 24 - 36 = 36 > 0$. So the function should be increasing at $x = 4$.

Thus, $f(x)$ is increasing for $x > -5$. Therefore, $A = (-5, \infty)$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when finding the derivative and solving inequalities.
• Interval Consideration: Always remember to consider the specific interval that each piece of the piecewise function is defined on. This is crucial for correctly interpreting the sign of the derivative.
• Endpoint Analysis: While not always necessary, it's good practice to check the endpoints of the intervals to see if they should be included or excluded based on the derivative's sign and the function's continuity.

Summary

We found the derivative of each piece of the piecewise function and determined the intervals where the derivative is greater than or equal to zero. This indicated where the function is increasing. Combining these intervals, we found that the function is increasing on $(-5, \infty)$.

Final Answer

The final answer is \boxed{(-5,\infty)}, which corresponds to option (A).