Find function $f'(x)$ such that its domain $D'=\mathbb{R}$, $f'(x)=f(x)$ $\forall x\in D$ and $f'(x)$ is continuous. Let $f(x)=\frac{^3\sqrt{x^3+3x^2+7}}{x+2}$.
I was asked to find $f'(x)$ such that 
$a)$ the domain $D'$ of $f'(x)$ is $\mathbb{R}$, 
$b)$ $f(x)=f'(x)$ $\forall x\in D$, with $D$ being domain of $f(x).$
$c)$ $f'(x)$ is continuous in $\mathbb{R}$.
I tried to define a function that $f'(x)$ that equals $f(x)$ when $x\neq-2$ and equals $c$ when $x=-2$, with $c$ being the number that makes the function continuous; this is, such that $\lim_{x\to-2} f(x) = c$. But I was enable to find the solution to this limit. Perhaps this is the way to go and I just didn't find a way to solve the limit, and perhaps my whole approach is wrong as a concept, I don't know. I was hoping someone could help me out with this problem.

We have
$$\lim_{x\to-2^+} \frac{\sqrt[3]{x^3+3x^2+7}}{x+2} = \frac{\sqrt[3]{11}}{0^+} = +\infty$$
so $f$ cannot be extended to a continuous function on $\mathbb{R} \to \mathbb{R}$.
