Systematic curve sketching with the derivative

A step-by-step routine for sketching curves: domain, symmetry, intercepts, asymptotes, stationary points and inflections, and using the sketch to solve polynomial inequations.

The routine

Not every step gives something for every curve, but checking them all stops you missing a feature.

Example 1: a cubic●●

Sketch $y = x^3 - 3x$.

Domain: all real $x$. Odd function, since $f(-x) = -f(x)$, so the curve has point symmetry about the origin.

Intercepts: $x(x^2 - 3) = 0$ gives $x = 0, \pm\sqrt3$.

$y' = 3x^2 - 3 = 0$ at $x = \pm 1$. $\;y'' = 6x$: $y''(-1) < 0$ and $y''(1) > 0$. So $(-1, 2)$ is a maximum and $(1, -2)$ is a minimum.

$y'' = 6x$ changes sign at $0$, so $(0, 0)$ is a point of inflection.

y = x³ − 3x

Odd, with zeroes at 0 and ±√3, turning points (−1, 2) and (1, −2), and an inflection at the origin.

Try this

Use the graph to solve $x^3 - 3x > 0$.

Example 2: solving an inequation●●

Solve $x^3 - 3x > 0$.

The curve is above the $x$-axis for $-\sqrt3 < x < 0$ or $x > \sqrt3$.

Example 3: a curve with an asymptote●●●

Sketch $y = \dfrac{1}{1 + x^2}$.

Domain: all real $x$ (the denominator is never $0$). Even, so symmetric in the $y$-axis. Always positive, $y$-intercept $1$.

As $x \to \pm\infty$, $y \to 0$: horizontal asymptote $y = 0$.

$y' = -\dfrac{2x}{(1 + x^2)^2}$, which is $0$ at $x = 0$, positive to the left and negative to the right: maximum $(0, 1)$.

$y'' = \dfrac{6x^2 - 2}{(1 + x^2)^3}$, which changes sign at $x = \pm\dfrac{1}{\sqrt3}$. Points of inflection at $\left(\pm\dfrac{1}{\sqrt3}, \dfrac34\right)$.

y = 1/(1 + x²)

A bell-shaped curve: maximum at (0, 1), inflections at x = ±1/√3, and asymptote y = 0.

Try this

Where is the curve concave down?

Flashcards

Sketching routine

Domain, symmetry, intercepts, asymptotes, $f'$, $f''$, then draw.

Odd function

$f(-x) = -f(x)$: point symmetry about the origin.

Solve $f(x) > 0$ from a sketch

Read off where the curve is above the $x$-axis.

Practice

Find the $y$-coordinate of the maximum turning point of $y = x^3 - 3x$.
How many $x$-intercepts does $y = x^3 - 3x$ have?
Solve $x^3 - 3x > 0$.
The curve $y = x^3 - 3x$ is:
Find the maximum value of $y = \dfrac{1}{1 + x^2}$.
Find the $y$-coordinate of each point of inflection of $y = \dfrac{1}{1 + x^2}$.
The horizontal asymptote of $y = \dfrac{1}{1 + x^2}$ is:
Which is usually the first step of the routine?
Vertical asymptotes of a rational function are found where: