Complete curve sketching - the six steps

Curve sketching is the systematic procedure for drawing the qualitative graph of a function, without a calculator, using only differential calculus. One follows a precise scheme in six steps.

In brief — The six steps of curve sketching

1. Domain of the function. (Where is it defined?)
2. Limits at the edges of the domain $\to$ asymptotes (horizontal, vertical, oblique).
3. Sign of the function: $f(x)>0$? $f(x)<0$? Where does it cross the $x$-axis?
4. First derivative: sign of $f'(x)$ $\to$ intervals of increase/decrease, relative maxima/minima. Absolute maxima/minima with the table of edges + stationary points.
5. Second derivative: sign of $f''(x)$ $\to$ concavity/convexity, inflection points.
6. Graph: assemble all the information into a drawing.

Each step adds a piece of information to the graph: the domain says “where” to draw, the limits and asymptotes describe the behaviour at the edges, the sign places the curve above or below the $x$-axis, the first derivative gives the “bumps and valleys” shape, the second derivative the curvature. The last step puts everything together.

Example — Complete study: $f(x) = \sin x\cdot e^{-\cos x}$

Step 1 — Domain: $f$ is defined for every $x\in\mathbb{R}$ (a product of functions defined everywhere). $f$ is periodic with period $2\pi$.

Step 2 — Symmetries: $f(-x) = \sin(-x)\cdot e^{-\cos(-x)} = -\sin x\cdot e^{-\cos x} = -f(x)$. Hence $f$ is odd (graph symmetric with respect to the origin). It is enough to study $f$ on $[0,2\pi]$ (or on $[0,\pi]$ and reconstruct by symmetry).

Step 3 — Sign: $e^{-\cos x}>0$ always. So $f(x)>0 \iff \sin x > 0 \iff x\in(0,\pi)$ (in the first period); $f(x)<0$ for $x\in(\pi,2\pi)$; $f(x)=0$ for $x=0, \pi, 2\pi$.

Sign of $f$ in the first period: positive on $(0,\pi)$, negative on $(\pi,2\pi)$.

Step 4 — First derivative: $f'(x) = \cos x\cdot e^{-\cos x} + \sin x\cdot e^{-\cos x}\cdot\sin x = e^{-\cos x}(\cos x + \sin^2 x).$ Since $e^{-\cos x}>0$, the sign of $f'$ is that of $\cos x + \sin^2 x = \cos x + 1 - \cos^2 x$. Setting $u=\cos x$ we study $-u^2+u+1 > 0$, i.e. $u^2-u-1<0$ with $u\in[-1,1]$. With $\Delta = 1+4 = 5$ we get $u = \dfrac{1\pm\sqrt{5}}{2}$; the only root in $[-1,1]$ is $u_0 = \dfrac{1-\sqrt{5}}{2}\approx -0{,}618$. Hence $f'(x)>0$ when $\cos x > u_0$; in $(0,2\pi)$ we have $\cos x = u_0$ for $x\approx 2{,}24$ (relative maximum) and $x\approx 4{,}04$ (relative minimum).

Step 5 — Second derivative: left as an exercise; the sign of $f''$ determines the inflection points.

Step 6 — Graph:

Graph of $f(x)=\sin x\cdot e^{-\cos x}$ on $[0,2\pi]$: relative maximum $\approx 1{,}46$ near $x\approx 2{,}24$, relative minimum $\approx -1{,}46$ near $x\approx 4{,}04$; by odd symmetry the graph extends about the origin.

Links

Topics: Curve sketching
Concepts: Asymptote · Concavity and convexity · Increase and decrease · Domain · Inflection point · Relative maxima and minima · Sign · Curve sketching
Methods: Curve sketching
Skills: Sketching a function · Drawing a graph