Graphing the reciprocal of a function (WACE Year 12 Mathematics Specialist Unit 3)

What this dot point is asking

SCSA wants you to transform a known graph into the graph of its reciprocal using these structural rules, without computing a new formula point by point.

Zeros become asymptotes

Wherever $f(x) = 0$, the reciprocal $\tfrac{1}{f(x)}$ is undefined and blows up, so each zero of $f$ gives a vertical asymptote of the reciprocal. The sign of $f$ just either side of the zero tells you whether the reciprocal branch goes to $+\infty$ or $-\infty$.

Large becomes small, small becomes large

A worked structural reading

Take a cubic $f$ with zeros at $x = -1, 1, 3$ and the usual cubic shape. The reciprocal $\dfrac{1}{f}$ then has three vertical asymptotes, one at each of $x = -1, 1, 3$. Between consecutive zeros $f$ changes sign, so the reciprocal also changes sign across each asymptote. Wherever $f$ has a local maximum or minimum, the reciprocal has a stationary point of the opposite type (provided $f$ is non-zero there), and far out where the cubic grows large the reciprocal hugs the $x$-axis. Reading the reciprocal off the original graph this way, feature by feature, is faster and less error-prone than substituting numbers, and it is exactly the reasoning SCSA expects in a sketch.

Sign and fixed points

The reciprocal has the same sign as $f$, since $\tfrac{1}{f}$ is positive exactly when $f$ is positive. The graphs cross where $f(x) = \tfrac{1}{f(x)}$, that is where $f(x) = \pm 1$; these points are shared by both curves.

Turning points swap type

Where $f$ has a local maximum (and is positive there), $\tfrac{1}{f}$ has a local minimum at the same $x$, because dividing one into a peak gives a trough. Likewise a positive local minimum of $f$ becomes a local maximum of the reciprocal. Where $f$ is negative the roles invert accordingly.

Why the turning-point swap happens

The swap of maxima and minima follows from calculus, which is worth knowing even though the sketch is done by reasoning. If $g(x) = \dfrac{1}{f(x)}$ then $g'(x) = -\dfrac{f'(x)}{[f(x)]^2}$. The denominator $[f(x)]^2$ is always positive, so $g'(x) = 0$ exactly where $f'(x) = 0$: the reciprocal has its stationary points at the same $x$-values as $f$. The sign of $g'$ is the opposite of the sign of $f'$, so the curve that was rising is now falling and vice versa, which is precisely why a maximum of a positive $f$ becomes a minimum of $\dfrac{1}{f}$.

Asymptotes of f become zeros of the reciprocal

The relationship runs both ways. Just as a zero of $f$ becomes a vertical asymptote of $\dfrac{1}{f}$, a vertical asymptote of $f$ becomes a point where $\dfrac{1}{f} \to 0$. So if $f$ itself is a rational function with a vertical asymptote at $x = a$, the reciprocal approaches the $x$-axis there. Tracking both directions of this duality lets you sketch the reciprocal of quite complicated graphs purely structurally.

Sketching routine

Mark the zeros of $f$ as asymptotes of the reciprocal, mark the $f = \pm 1$ crossing points, send the reciprocal toward $y = 0$ where $f$ is large, and turn the peaks into troughs. Then join smoothly.