Vertical tangent

Vertical tangent on the function ƒ(x) at x = c.

In mathematics, particularly calculus, a vertical tangent is a tangent line that is vertical. Because a vertical line has infinite slope, a function whose graph has a vertical tangent is not differentiable at the point of tangency.

Limit definition

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

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

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

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

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

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

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.

Vertical cusps

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

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

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

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

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.

Example

The function

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

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

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

Similarly, the function

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

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

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

and

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

References

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