Vertical tangent

A function ƒ has a vertical tangent at x = a if the difference quotient used to define the derivative has infinite limit:

${\displaystyle \lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}={+\infty }\quad {\text{or}}\quad \lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}={-\infty }.}$ 

The graph of ƒ has a vertical tangent at x = a if the derivative of ƒ at a is either positive or negative infinity.

For a continuous function, it is often possible to detect a vertical tangent by taking the limit of the derivative. If

${\displaystyle \lim _{x\to a}f'(x)={+\infty }{\text{,}}}$ 

then ƒ must have an upward-sloping vertical tangent at x = a. Similarly, if

${\displaystyle \lim _{x\to a}f'(x)={-\infty }{\text{,}}}$ 

then ƒ must have a downward-sloping vertical tangent at x = a. In these situations, the vertical tangent to ƒ appears as a vertical asymptote on the graph of the derivative.

Closely related to vertical tangents are vertical cusps. This occurs when the one-sided derivatives are both infinite, but one is positive and the other is negative. For example, if

${\displaystyle \lim _{h\to 0^{-}}{\frac {f(a+h)-f(a)}{h}}={+\infty }\quad {\text{and}}\quad \lim _{h\to 0^{+}}{\frac {f(a+h)-f(a)}{h}}={-\infty }{\text{,}}}$ 

then the graph of ƒ will have a vertical cusp that slopes up on the left side and down on the right side.

As with vertical tangents, vertical cusps can sometimes be detected for a continuous function by examining the limit of the derivative. For example, if

${\displaystyle \lim _{x\to a^{-}}f'(x)={-\infty }\quad {\text{and}}\quad \lim _{x\to a^{+}}f'(x)={+\infty }{\text{,}}}$ 

then the graph of ƒ will have a vertical cusp at x = a that slopes down on the left side and up on the right side.

The function

${\displaystyle f(x)={\sqrt[{3}]{x}}}$ 

has a vertical tangent at x = 0, since it is continuous and

${\displaystyle \lim _{x\to 0}f'(x)\;=\;\lim _{x\to 0}{\frac {1}{3{\sqrt[{3}]{x^{2}}}}}\;=\;\infty .}$ 

Similarly, the function

${\displaystyle g(x)={\sqrt[{3}]{x^{2}}}}$ 

has a vertical cusp at x = 0, since it is continuous,

${\displaystyle \lim _{x\to 0^{-}}g'(x)\;=\;\lim _{x\to 0^{-}}{\frac {2}{3{\sqrt[{3}]{x}}}}\;=\;{-\infty }{\text{,}}}$ 

and

${\displaystyle \lim _{x\to 0^{+}}g'(x)\;=\;\lim _{x\to 0^{+}}{\frac {2}{3{\sqrt[{3}]{x}}}}\;=\;{+\infty }{\text{.}}}$ 

• Vertical Tangents and Cusps. Retrieved May 12, 2006.