Curve sketching | Calculus

The checklist

Calculus lets you sketch an accurate graph without plotting dozens of points. Gather the information in this order:

A complete example: $f(x) = \frac{x}{x^2 + 1}$

1. Domain. $$x^2 + 1$$ is never zero: all real numbers.

2. Intercepts. $$f(0) = 0$$ , and $$f(x) = 0$$ only at $$x = 0$$ . The only intercept is the origin.

3. Symmetry. $f(-x) = \frac{-x}{x^2 + 1} = -f(x)$ : odd, symmetric about the origin.

4. Asymptotes. No vertical asymptotes. The denominator has the larger degree, so $\lim_{x \to \pm\infty} f(x) = 0$ : horizontal asymptote $$y = 0$$ .

5. First derivative.

f'(x) = \frac{(x^2 + 1) - x \cdot 2x}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}

The sign is that of $$1 - x^2$$ : negative for $$x < -1$$ , positive on $$(-1, 1)$$ , negative for $$x > 1$$ .

Relative minimum at $\left(-1, -\frac{1}{2}\right)$ ; relative maximum at $\left(1, \frac{1}{2}\right)$ .

6. Second derivative.

f''(x) = \frac{2x^3 - 6x}{(x^2 + 1)^3} = \frac{2x(x^2 - 3)}{(x^2 + 1)^3}

It changes sign at $x = -\sqrt{3}$ , $$0$$ and $\sqrt{3}$ :

	$x < -\sqrt{3}$	$-\sqrt{3} < x < 0$	$0 < x < \sqrt{3}$	$x > \sqrt{3}$
$$f''(x)$$	$$-$$	$$+$$	$$-$$	$$+$$
concavity	down	up	down	up

Inflection points at $\left(-\sqrt{3}, -\frac{\sqrt{3}}{4}\right)$ , $$(0, 0)$$ and $\left(\sqrt{3}, \frac{\sqrt{3}}{4}\right)$ .

7. Sketch. On the far left the graph starts just below the $$x$$ -axis, dips to the minimum at $\left(-1, -\frac{1}{2}\right)$ , climbs through the origin to the maximum at $\left(1, \frac{1}{2}\right)$ , then falls back toward the $$x$$ -axis, staying just above it. The S-shape is symmetric about the origin, and the range is $\left[-\frac{1}{2}, \frac{1}{2}\right]$ .

Practice

Quick check

Which step of curve sketching tells you where the graph is concave up?