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Corner points, cusps, points with vertical/horizontal tangent

Part 4 · Derivatives · Chapter 3 · lecture notes by Fabio Furini · Chapter PDF

1. Right derivative and left derivative

Definition 1: of right (left) derivative

Let \(f: (a, b) \rr \R\); the function \(f\) is said to be differentiable at \(x_0 \in (a, b)\) from the right (from the left) if the following limit exists and is finite

\[ \lim_{h \rr 0^+} \frac{f(x_0 + h) - f(x_0)}{h} \quad \left(~ \lim_{h \rr 0^-} \frac{f(x_0 + h) - f(x_0)}{h} ~\right) \]

then \(f\) is differentiable from the right (from the left) and the limit is called the right derivative (left derivative).

  • The right derivative is denoted by the symbol \(f'_{+}(x_0)\), while the left derivative is denoted by the symbol \(f'_{-}(x_0)\)

2. Corner points

Definition 2: of corner point

If \(f\) is continuous and differentiable from the right and from the left (but not differentiable) at \(x_0\), we say that \(f\) has a corner point at \(x = x_0\).

Example 1: Corner points

Consider the function \(f(x) = |x|\) and the point \(x_0=0\). We have:

\[ f'_{+}(0)=\lim_{h \rr 0^+}\frac{|h|}{h} = \lim_{h \rr 0^+} \frac{h}{h} = 1 {\rm ~~~~and~~~~} f'_{-}(0)=\lim_{h \rr 0^-} \frac{|h|}{h} = \lim_{h \rr 0^-} \frac{-h}{h} = -1 \]

Since the limit of the difference quotient does not exist, \(f\) is not differentiable at \(x =0\). The function is continuous at \(x=0\) (the origin) since

\[ \lim_{x \rr 0^+} f(x)=\lim_{x \rr 0^+} x = 0 {\rm ~~~and~~~} \lim_{x \rr 0^-} f(x) =\lim_{x \rr 0^-} -x = 0 {\rm ~~~~hence~~~} \lim_{x \rr 0} f(x) = f(0)=0 \]

Since the right and left limits of the difference quotient at \(x=0\) exist and are finite, the graph therefore has a corner point at \(x=0\).

Figure 1

The formula that concisely expresses the derivative of the absolute value function (away from the origin) is the following:

\[ f(x)= |x|, \quad f'(x) = sgn (x) = \begin{cases} 1 & {\rm if~~} x > 0\\ -1 & {\rm if~~} x < 0 \end{cases} \]

3. Points with vertical/horizontal tangent

  • If \(f\) is continuous at a point \(x_0\) and

    \[ \lim_{h \rr 0} \frac{f(x_0 + h) - f(x_0)}{h}= \pm \infty \]

    then \(f\) is not differentiable at \(x_0\) but, geometrically, the graph of \(f\) has a well-defined tangent line parallel to the \(y\)-axis.

    In this case we will allow the notation

    \[ f'(x_0) = \ip, ~~~~ f'(x_0) = \im \]

    and we will speak of a point with vertical tangent.

Definition 3: of point with vertical tangent

If

\[ f'(x_0) = \ip {\rm ~~~or~~~} f'(x_0) = \im \]

we say that \(f\) has a point with vertical tangent at \(x = x_0\).

  • Similarly, if \(f\) is continuous at a point \(x_0\) and

    \[ \lim_{h \rr 0} \frac{f(x_0 + h) - f(x_0)}{h}= 0 \]

    the graph of \(f\) has a well-defined tangent line parallel to the \(x\)-axis. In this case we will speak of a point with horizontal tangent.

Definition 4: of point with horizontal tangent

If

\[ f'(x_0) = 0 \]

we say that \(f\) has a point with horizontal tangent at \(x = x_0\).

Example 2: Point with vertical tangent

Let

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

then for \(x_0=0\)

\[ \frac{f(h) - f(0)}{h} = \frac{ \sqrt[3]{h}}{h} = \frac{ 1}{h^{2/3}} \]

and the limits are:

\[ \lim_{h \rr 0^-} \frac{ 1}{h^{2/3}} = \ip {\rm ~~~and~~~}\lim_{h \rr 0^+} \frac{ 1}{h^{2/3}} = \ip {\rm ~~~~hence~~~~} \lim_{h \rr 0} \frac{ 1}{h^{2/3}} = \ip {\rm ~~~and~~~} f'(0)= \ip. \]

Figure 2

The function has a point with vertical tangent at \(x_0=0\). The function:

\[ f(h) = \frac{1}{h^{{2}/{3}}} = h^{-\frac{2}{3}} \]

is a power with negative rational exponent \(\frac{m}{n}\) with \(n\) odd (hence defined on \(\R \setminus \{0\}\)) and \(m\) even (hence an even function). Its graph is:

Figure 3

Example 3: Point with vertical tangent

Let

\[ f(x) = -\sqrt[3]{x-2}+1 \]

then for \(x_0=2\)

\[ \frac{f(2 + h) - f(2)}{h} = \frac{- \sqrt[3]{(2+h)-2}+1 - 1}{h} = -\frac{ \sqrt[3]{h}}{h}=-\frac{ 1}{h^{2/3}} \]

and the limits are

\[ \lim_{h \rr 0^-} -\frac{ 1}{h^{2/3}} = \im {\rm ~~~and~~~}\lim_{h \rr 0^+} -\frac{ 1}{h^{2/3}} = \im {\rm ~~~~hence~~~~} \lim_{h \rr 0} - \frac{ 1}{h^{2/3}} = \im {\rm ~~~and~~~} f'(2)= \im. \]

Figure 4

The function has a point with vertical tangent at \(x_0=2\).

4. Cusps

Definition 5: of cusp

Let \(f\) be a function continuous at \(x_0\); if

\[ f'_+(x_0)= \pm \infty {\rm ~~and~~} f'_-(x_0)= \mp \infty \]

we say that \(f\) has a cusp at \(x_0\).

Example 4: Cusp

Let

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

then for \(x_0=0\)

\[ \frac{f(h) - f(0)}{h} = \frac{ \sqrt[3]{|h|}}{h} \]

and the limits are:

\[ \lim_{h \rr 0^-} \frac{ \sqrt[3]{-h}}{h}=\lim_{h \rr 0^-} -\frac{ 1}{h^{2/3}} = \im {\rm ~~~~and~~~~}\lim_{h \rr 0^+} \frac{ \sqrt[3]{h}}{h} =\lim_{h \rr 0^+} \frac{ 1}{h^{2/3}} = \ip, \]

hence:

\[ f'_-(0)= \im {\rm ~~~~and~~~~}f'_+(0)= \ip. \]

Figure 5

The function has a cusp at \(x_0=0\).

  • In the mixed case where one of the two derivatives is finite and the other is infinite (with \(f\) continuous) we still speak of a corner point.

  • Finally, if the function is defined only for \(x \ge x_0\) and at that point has an infinite (right) derivative, we will simply say that at that point it has a vertical tangent, without speaking of either a cusp or a point with vertical tangent.

Example 5: Point with vertical tangent

Let

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

then for \(x_0=0\)

\[ \frac{f(h) - f(0)}{h} = \frac{ \sqrt{h}}{h} = \frac{1}{h^{1/2}} \]

we only have the right limit

\[ \lim_{h \rr 0^+} \frac{1}{h^{1/2}} = \ip {\rm ~~~and~~~}f'_+(0)= \ip. \]

Figure 6

The function has a point with vertical tangent at \(x_0=0\). The function:

\[ f(h) = \frac{1}{h^{{1}/{2}}} = h^{-\frac{1}{2}} \]

is a power with negative rational exponent \(\frac{m}{n}\) with \(n\) even (hence defined only on \(\R_+\)). Its graph is:

Figure 7

Example 6: Continuous extension from the right and behavior at the origin

Let

\[ f(x) = x\: \log x, {\rm ~~for~~} x>0 {\rm ~~we~have~~} \lim_{x \rr 0^+} x\: \log x = 0 \]

hence the function can be extended by continuity from the right at \(x = 0\), by setting:

\[ f(x)= \begin{cases} x\: \log x & {\rm if~~} x > 0\\[2ex] 0 & {\rm if~~} x = 0 \end{cases} \]

Figure 8

We compute the right derivative at \(x_0=0\) of \(f(x) = x\: \log x\) extended by continuity from the right at \(x=0\). We have, for \(x_0=0\):

\[ \frac{f(x_0 + h) - \overbrace{f(x_0)}^{=0}}{h}=\frac{f(x_0 + h)}{h} = \frac{f(h)}{h} \]

and consequently

\[ \lim_{h \rr 0^+}\frac{f(x_0 + h) - f(x_0)}{h}= \lim_{h \rr 0^+}\frac{h \: \log h}{h}= \lim_{h \rr 0^+} \log h = \im {\rm ~~~~hence~~~~} f'_+(x_0)= \im \]

The function therefore has a point with vertical tangent at \(x_0=0\).

Example 7: Continuous extension from the right and behavior at the origin

Let

\[ f(x) = e^{-\frac{1}{x}}, {\rm ~~for~~} x>0 {\rm ~~we~have~~} \lim_{x \rr 0^+} e^{-\frac{1}{x}} = 0 \]

hence the function can be extended by continuity from the right at \(x = 0\), by setting:

\[ f(x)= \begin{cases} e^{-\frac{1}{x}} & {\rm if~~} x > 0\\[2ex] 0 & {\rm if~~} x = 0 \end{cases} \]

Figure 9

We compute the right derivative at \(x_0=0\) of \(f(x) = e^{-\frac{1}{x}}\) extended by continuity from the right at \(x=0\).

We have:

\[ \lim_{h \rr 0^+}\frac{f(x_0 + h) - f(x_0)}{h}= \lim_{h \rr 0^+}\frac{e^{-\frac{1}{h}}}{h}= \lim_{h \rr 0^+} \frac{1}{e^{\frac{1}{h}} \; h} \]

Changing the variable:

\[ y=\frac{1}{h}, {\rm ~~if~~} h \rr 0^{+} {\rm ~~then~~} y \rr \ip \]

We obtain:

\[ \lim_{h \rr 0^+} \frac{1}{e^{\frac{1}{h}} \; h} = \lim_{y \rr \ip} \frac{y}{e^y} = 0 {\rm ~~~~hence~~~~} f'_+(x_0)= 0 \]

and the function has a point with horizontal tangent (from the right) at \(x_0=0\).