1. Given piecewise function

We need the set

$A=\{x\in \mathbb R: f \text{ is increasing at } x\}.$

We check monotonicity on each interval using derivatives.

2. For $x<-5$

Here,

$f(x)=-55x$

So,

$f'(x)=-55<0$

Hence $f$ is decreasing on $(-\infty,-5)$.

So no point from this interval belongs to $A$.

3. For $-5\le x\le 4$

Here,

$f(x)=2x^3-3x^2-120x$

Differentiate:

$f'(x)=6x^2-6x-120=6(x^2-x-20)=6(x-5)(x+4)$

Now on the interval $[-5,4]$, the sign depends on $(x-5)(x+4)$.

Critical points are $x=-4$ and $x=5$, but only $x=-4$ lies in $[-5,4]$.

Sign analysis in $[-5,4]$:

• For $-5<x<-4$: both $(x-5)<0$ and $(x+4)<0$, so $f'(x)>0$
• At $x=-4$: $f'(-4)=0$
• For $-4<x<4$: $(x-5)<0$, $(x+4)>0$, so $f'(x)<0$

Thus $f$ is increasing on

$(-5,-4)$

4. For $x>4$

Here,

$f(x)=2x^3-3x^2-36x-336$

Differentiate:

$f'(x)=6x^2-6x-36=6(x^2-x-6)=6(x-3)(x+2)$

For $x>4$, we have:

• $x-3>0$
• $x+2>0$

Hence,

$f'(x)>0 \quad \text{for all } x>4$

So $f$ is increasing on

$(4,\infty)$

5. Check junction points $x=-5$ and $x=4$

Since options are given in open intervals, let us see whether endpoints are included in the increasing set.

• At $x=-5$, the function changes from decreasing on the left to increasing on the right, so $x=-5$ is not included in an interval of increase.
• At $x=4$, the function is decreasing just to the left and increasing just to the right, so $x=4$ is not included either.

Thus,

$A=(-5,-4)\cup(4,\infty)$

6. Compare with options

This matches Option C.

$\boxed{A=(-5,-4)\cup(4,\infty)}$