Q:

Vertical cusp vertical tangent question

Can anyone help me solve the following two problems:

Does the following function have a vertical  tangent or vertical cusp at $c=0$

$$f(x)=(3+x^{\frac{2}{5}})$$

I got for the derivative
$$f'(x)=\frac{2}{5\sqrt[5]{x}^3}$$
Now I think this would be a cusp because as x approaches zero from the left and right f(x) approches infinity.

My second question is with the function $f(x)=|{(x+8)^{1/3}}|$.

For my derivative I got
$$f'(x)=\frac{1}{3}(x+8)^{\frac{-2}{3}},\;x > -8$$
$$f'(x)=\frac{-1}{3}(x+8)^{\frac{-2}{3}},\;x < -8$$
Would this not be a tangent because of the square expoent.

A:

Yes, your first function has a cusp at $x = 0$, because the derivative $f'(x)$ approaches $\infty$ as $x \to 0^+$ and as $x\to 0^-$. We can see this at the point $(0, 3)$:
$\quad f(x)=\left(3+x^{\large\frac{2}{5}}\right)$ (see blue curve)

Your second function also has a cusp: at $x = -8$
Can you try to rewrite your derivative, or redefine if you mean "for $x \geq 0, f'(x) = ...$ and for $x < 0, f'(x) = ...?$. 
$\quad f(x)=|(x+8)^{1/3}|$

