Side limits of the derivative of this function $f:\mathbb{R}\to \mathbb{R}$ with $f\left(x\right)=\left(x^3+3x^2-4\right)^{\frac{1}{3}}$
Calculate side limits of this function's derivative, $f'_s\:and\:f'_d$, in $x_o=-2$
The answer key says I should get $\infty $ and $-\infty$ but I'm not getting that. The derivative I get is $\frac{x\left(x+2\right)}{\left(\left(x-2\right)^2\left(x+2\right)^4\right)^{\frac{1}{3}}}$ and by doing the multiplication from the denominator I would get something with $x^2$.
 A: You have the wrong expression for the derivative.
$$\begin{align}
\frac d{dx}\left[(x^3+3x^2-4)^{1/3}\right]
  &= \frac 13(x^3+3x^2-4)^{-2/3}(3x^2+6x) \\[2ex]
  &= \frac {x^2+2x}{[(x+2)^2(x-1)]^{2/3}} \\[2ex]
  &= \frac {x(x+2)}{(x+2)^{4/3}(x-1)^{2/3}} \\[2ex]
  &= \frac {x}{(x+2)^{1/3}(x-1)^{2/3}} \\[2ex]
\end{align}$$
That last expression's denominator tends to zero as $x\to-2$ but the numerator does not tend to zero, which means an infinite limit on both sides of $-2$. As $x\to-2$ from the left, both numerator and denominator are negative, so the expression tends to $+\infty$. As $x\to-2$ from the right, the numerator is negative but the denominator is positive, so the expression tends to $-\infty$.
A: It might help to observe that $f(x) \; = \; (x+2)^{2/3}(x-1)^{1/3}.$ To obtain this factorization, note that $x^3 + 3x^2 - 2$ equals zero when $x = -2,$ so we know $x+2$ is a factor of $x^3 + 3x^2 - 2.$ Use long division, or use synthetic division, or note that $x^3 + 3x^2 - 2 = x^3 + 2x^2 + x^2 - 2$ (and factor by grouping), and you'll find that $x^3 + 3x^2 - 2 = (x+2)(x^2 + x - 2).$ Now factor the quadratic. Since this "looks like" the graph of $y = -3^{1/3}(x+2)^{2/3}$ when $x$ is close to $-2,$ the graph will look like that of $y = x^{2/3}$ translated left $2$ units and reflected about the $x$-axis and slightly stretched, and thus we'd expect the left and right limits of the derivative to be what you said the answer is. However, two graphs can be very close to each other and still have vastly different derivative behaviors. Consider, for example, the graph of $y = x^2$ and the graph of $y = x^2 + 10^{-100}W(x),$ where $W(x)$ is the Weierstrass nowhere differentiable function. So let's carry out these limits explicitly.
Assuming $x \neq -2$ and using the product rule, we get
$$f'(x) \;\; = \;\; \frac{2}{3}(x+2)^{-\frac{1}{3}}(x-1)^{\frac{1}{3}} \;\; + \;\; \frac{1}{3}(x+2)^{\frac{2}{3}}(x-1)^{-\frac{2}{3}}$$
The two limits are now straightforward.
For $x \rightarrow -2$ from the left we get
$$f'(x) \;\; \longrightarrow \;\; \frac{2}{3}(\rightarrow -\infty)(\rightarrow -3^{\frac{1}{3}}) \;\; + \;\; \frac{1}{3}(\rightarrow 0)(\rightarrow 3^{\frac{2}{3}}) \;\; = \;\; +\infty $$
and for $x \rightarrow -2$ from the right we get
$$f'(x) \;\; \longrightarrow \;\; \frac{2}{3}(\rightarrow +\infty)(\rightarrow -3^{\frac{1}{3}}) \;\; + \;\; \frac{1}{3}(\rightarrow 0)(\rightarrow 3^{\frac{2}{3}}) \;\; = \;\; -\infty $$
