The Graph of y = 1/f(x)

One height at a time

To sketch $y=\frac{1}{f\left(x\right)}$y = 1/f(x) from y = f(x), take each x and replace the height f(x) by its reciprocal. The x-values stay where they are; only the heights change.

Where f(x) = 0 there is no reciprocal, so $\frac{1}{f\left(x\right)}$1/f(x) has no value there. As x approaches such a point, f(x) gets close to 0 and $\frac{1}{f\left(x\right)}$1/f(x) grows without limit. For $f\left(x\right)=x^2-4=\left(x+2\right)\left(x-2\right)$f(x) = x² − 4 = (x + 2)(x − 2), the zeros at x = −2 and x = 2 become vertical asymptotes of $y=\frac{1}{f\left(x\right)}$y = 1/f(x).

Fixed points and signs

A number equals its own reciprocal only when it is 1 or −1. So wherever f(x) = 1 or f(x) = −1, the two graphs meet. Here $x^2-4=1$x² − 4 = 1 at $x=\pm \sqrt{5}\approx \pm 2.236$x = ±√5 ≈ ±2.236, and $x^2-4=-1$x² − 4 = −1 at $x=\pm \sqrt{3}\approx \pm 1.732$x = ±√3 ≈ ±1.732.

A reciprocal has the same sign as the number, so $\frac{1}{f\left(x\right)}$1/f(x) is above the x-axis exactly where f(x) is. Here that is for x < −2 and x > 2; between −2 and 2 both graphs are below the axis.

Large and small swap

Where f(x) is large, $\frac{1}{f\left(x\right)}$1/f(x) is small, and where f(x) is small, $\frac{1}{f\left(x\right)}$1/f(x) is large. At x = 4, f = 12 and $\frac{1}{f}=\frac{1}{12}$1/f = 1/12; at x = 2.1, f = 0.41 and $\frac{1}{f}\approx 2.44$1/f ≈ 2.44.

As x runs far to the left or right, $x^2-4$x² − 4 grows without limit, so $\frac{1}{f\left(x\right)}$1/f(x) shrinks toward 0 from above: the x-axis is a horizontal asymptote.

Turning points

At a turning point of f that is not on the axis, $\frac{1}{f}$1/f has a turning point at the same x, of the opposite kind. $f\left(x\right)=x^2-4$f(x) = x² − 4 has its minimum at (0, −4); between the asymptotes the reciprocal is closest to 0 there, so $\frac{1}{f}$1/f has a maximum at $\left(0,-\frac{1}{4}\right)$(0, −1/4).

Its height is just 1 ÷ (−4). The turning point never moves sideways.

The gold curve is $y=\frac{1}{x^2-4}$y = 1/(x² − 4), in three pieces split by the asymptotes x = −2 and x = 2. The faint curve is $y=x^2-4$y = x² − 4, and the dots mark where the two meet, at heights 1 and −1.

When f has no zeros

Take $f\left(x\right)=x^2+1$f(x) = x² + 1. It is never 0, so $y=\frac{1}{x^2+1}$y = 1/(x² + 1) has no vertical asymptote and is one unbroken curve. f has its minimum at (0, 1), so the reciprocal has its maximum at $\left(0,\frac{1}{1}\right)=\left(0,1\right)$(0, 1/1) = (0, 1), where the two graphs meet.

Everywhere else $x^2+1>1$x² + 1 > 1, so the reciprocal is between 0 and 1, and it falls toward 0 as x runs away in either direction.

The gold curve $y=\frac{1}{x^2+1}$y = 1/(x² + 1) stays between 0 and 1 and peaks at (0, 1), where it touches the faint curve $y=x^2+1$y = x² + 1 at that curve's minimum.

The usual slips

Reciprocating the x-values: the zeros of f stay at the same x and become vertical asymptotes there.

Keeping a minimum as a minimum: where f is least in size, its reciprocal is greatest in size.

Putting a meeting point where f(x) = 0: there the reciprocal does not exist.