So there are 3 situations that a function is not differentiable 1. a vertical tangent, 2. a discontinuity and 3. at a cusp. Consider the function $f(x)=e^{-|x|}$ at the point $x=0$. It is continuous here but its first derivative is not does, and therefore it cannot be differential here can it? But I have read that a cusp is when the limit of the first derivative must tend to $+\infty$ when approaching the point from one direction and $-\infty$ when from the other. In this case it tends to $+1$ and $-1$ which should mean that it does not have a cusp here and does not fall into one of the non differentiable categories. But it is not differentiable, so is this definition of a cusp wrong, if so what is the actual definition? thanks.
[Math] the definition of a cusp
derivativesgraphing-functions
Related Solutions
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 $x < 0, f'(x) = ...?$.
$\quad f(x)=|(x+8)^{1/3}|$
The definition of a vertical cusp is that the one-sided limits of the derivative approach opposite $ \pm \infty $: positive infinity on one side and negative infinity on the other side. A vertical tangent has the one-sided limits of the derivative equal to the same sign of infinity. As a result, the derivative at the relevant point is undefined in both the cusp and the vertical tangent.
You have a case where the derivative exists, as you showed in your question. Therefore, it is neither a cusp nor a vertical tangent.
Best Answer
Maxima or minima for functions occur at three types of points:
a. Stationary points where the derivative equals zero.
b. Corners where the left sided derivative does not equal the right sided derivative - for example, $y = |x|$. This is a perfect example of a corner.
c. Cusps, which are infinitely sharp corners. A cusp is a point where you have a vertical tangent, but with the following property: on one side the derivative is $+\infty$, on the other side the derivative is $-\infty$. The paradigm example was stated above: $y = x^{2\over3}$. The limit of the derivative as you approach zero from the left goes to $-\infty$. The limit of the derivative as you approach zero from the right goes to $+\infty$. At $x = 0$, there is a vertical tangent.
Vertical tangents exist at cusps (when describing functions). But a vertical tangent itself is not a cusp. The curve $y = x^{1\over3}$ has a vertical tangent at the origin but the there is no cusp: the curve is smooth, there is no pointy needle-like behavior to it.