Some less familiar curves

Using the derivative on rational functions with asymptotes and discontinuities, and recognising vertical tangents and cusps where the derivative is undefined.

Rational functions

Example 1●●●

Find the stationary points of $y = \dfrac{x}{x^2 + 1}$.

$y' = \dfrac{(x^2 + 1) - x \cdot 2x}{(x^2 + 1)^2} = \dfrac{1 - x^2}{(x^2 + 1)^2}$

Quotient rule with u = x and v = x² + 1.

$y' = 0$ when $x = \pm 1$. The denominator is always positive, so the sign of $y'$ is the sign of $1 - x^2$.

$y'$ goes $-, 0, +$ at $x = -1$ and $+, 0, -$ at $x = 1$. So $\left(-1, -\tfrac12\right)$ is a minimum and $\left(1, \tfrac12\right)$ is a maximum.

y = x/(x² + 1)

As x gets large, y gets close to 0, so the x-axis is an asymptote.

Try this

Why can the curve never go above $\tfrac12$?

Discontinuities in the table of slopes

For $y = \dfrac{1}{x - 2}$, $\;y' = -\dfrac{1}{(x - 2)^2}$ is negative everywhere it exists. The curve is decreasing on each side of the asymptote $x = 2$, but it is not decreasing "everywhere": $y(1) = -1$ is less than $y(3) = 1$.

Vertical tangents and cusps

Vertical tangent. $y = x^{\frac13}$ has $y' = \dfrac{1}{3x^{\frac23}}$, which is undefined at $x = 0$ and becomes huge nearby. The curve is smooth there but its tangent is vertical.

Cusp. $y = x^{\frac23}$ has $y' = \dfrac{2}{3x^{\frac13}}$, which changes from very negative to very positive at $x = 0$. The curve comes to a sharp point.

A vertical tangent and a cusp

Orange: y = x^(1/3), with a vertical tangent at the origin. Blue: y = x^(2/3), with a cusp at the origin.

Try this

Which curve is differentiable at $x = 0$? (Neither: $y'$ is undefined there for both.)

Flashcards

Stationary points of $dfrac{x}{x^2 + 1}$

Minimum $\left(-1, -\tfrac12\right)$, maximum $\left(1, \tfrac12\right)$.

Vertical tangent

$f'$ is undefined and the curve is smooth, with a vertical tangent.

Cusp

$f'$ is undefined and jumps from $-\infty$ to $+\infty$ (or back): a sharp point.

Practice

Find the maximum value of $y = \dfrac{x}{x^2 + 1}$.
Find the minimum value of $y = \dfrac{x}{x^2 + 1}$.
For $y = \dfrac{x}{x^2 + 1}$, find $y'(0)$.
The derivative of $\dfrac{x}{x^2 + 1}$ is:
At $x = 0$, the curve $y = x^{\frac13}$ has:
At $x = 0$, the curve $y = x^{\frac23}$ has:
For $y = \dfrac{1}{x - 2}$, which statement is true?
A table of slopes must include:
How many stationary points does $y = \dfrac{x}{x^2 + 1}$ have?