I have two questions regarding vertical tangent and vertical cups.
The first one is determine whether the graph of $f(x)=(x+5)^{\frac{-2}{7}}$ has a vertical tangent or vertical cusp $c=-5$
I personally would say no because it is undefined at $c=-5$
My second questions is
Determine whether the graph of $f(x)=(x-4)^{\frac{3}{7}}$ has a vertical tangent of vertical cusp at $c=4$
taking the derivative I got
$f(x)=\frac{3}{7}(x-4)^{\frac{-4}{7}}$
I would say this is is a vertical tangent because of the 4 exponent the limit of both left and right limits of c approach infinity.
Best Answer
Your first function does indeed have a vertical tangent at $x = -5$. See the graph of your function, below: blue curve (Wolfram Alpha).
For your first function: $$f(x) = \frac{1}{\sqrt[\large 7]{(x + 5)^2}}$$ $$f'(x) = -\frac{2}{7\sqrt[\large 7]{(x+5)^9}}$$
Note that the slope of the line tangent to the curve as $x \to -5^+$ DNE: the tangent becomes vertical because $f'(x) \to -\infty$, and the slope of the tangent line to the curve as $x\to -5^-$ DNE: the tangent becomes vertical because $f'(x) \to +\infty$.
In terms of testing for vertical tangency, a vertical tangent exists at a point $c$ whenever the limit of $f'(x) \to \infty\;$ and/or $\;f'(x) \to -\infty$ when approaching $c$ from the right, as well as from the left. The limits as one approaches a point of vertical tangenccy can both approach $+\infty$, both approach $-\infty$, or approach $+\infty$ from one direction and $-\infty$ from the other direction.
You're very much correct about your second function, there is a cusp at $x = 4$: for the very reasons you give.
See it as Wolfram plots the curve (blue):
$f(x) = (x-4)^{3/7}$.