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(D) To find where $f(x)$ is increasing,we calculate $f'(x)$ for each interval:$1$. For $x < -5$,$f(x) = -55x$,so $f'(x) = -55$. Since $f'(x) < 0$,the

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

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(D) To find where $f(x)$ is increasing,we calculate $f'(x)$ for each interval:$1$. For $x $2$. For $-5 For $f'(x) > 0$,we need $(x - 5)(x + 4) > 0$,which occurs when $x  5$. Within the interval $(-5, 4)$,this is satisfied for $x \in (-5, -4)$.$3$. For $x > 4$,$f(x) = 2x^3 - 3x^2 - 36x - 336$,so $f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x - 3)(x + 2)$.For $x > 4$,both $(x - 3)$ and $(x + 2)$ are positive,so $f'(x) > 0$ for all $x > 4$.Combining these,$f(x)$ is increasing on $(-5, -4) \cup (4, \infty)$.

Let $f : R \rightarrow R$ be defined as,
$f(x)=\begin{cases}-55 x, & \text{if } x<-5 \\ 2 x^{3}-3 x^{2}-120 x, & \text{if } -5 \leq x \leq 4 \\ 2 x^{3}-3 x^{2}-36 x-336, & \text{if } x>4 \end{cases}$
Let $A=\{ x \in R : f \text{ is increasing} \}$. Then $A$ is equal to :

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Let $f : R \rightarrow R$ be defined as,$f(x)=\begin{cases}-55 x, & \text{if } x<-5 \\ 2 x^{3}-3 x^{2}-120 x, & \text{if } -5 \leq x \leq 4 \\ 2 x^{3}-3 x^{2}-36 x-336, & \text{if } x>4 \end{cases}$Let $A=\{ x \in R : f \text{ is increasing} \}$. Then $A$ is equal to :