NSW HSC Mathematics Extension 1

The graph of the reciprocal, y = 1/f(x)

The graph of $y = \dfrac{1}{f(x)}$ can be sketched from the graph of $y = f(x)$ without any calculation, by reading off a few features of $f$.

The two graphs have the same sign everywhere, so the reciprocal never crosses the $x$-axis. The only points where the graphs meet are where $f(x)^2 = 1$, because $\dfrac{1}{f(x)} = f(x)$ rearranges to $f(x)^2 = 1$.

Where $f$ is positive and increasing, its reciprocal is positive and decreasing, and where $f$ is negative and increasing, its reciprocal is also decreasing. A maximum of $f$ at a positive value becomes a minimum of the reciprocal, and a minimum of $f$ at a positive value becomes a maximum.

Worked example

Sketch $y = \dfrac{1}{x^2 - 4}$ from $f(x) = x^2 - 4$.

The zeroes of $f$ are $x = \pm 2$, so the reciprocal has vertical asymptotes at $x = \pm 2$. As $x \to \pm\infty$, $f(x) \to \infty$, so $y \to 0$ from above. The function $f$ has a minimum value of $-4$ at $x = 0$, so the reciprocal has a turning point at $\left(0, -\dfrac14\right)$, and because $f$ is negative between the asymptotes, this is a local maximum of the reciprocal. For $|x| > 2$ the reciprocal is positive and decreasing towards $0$ as $|x|$ grows. The graphs meet where $x^2 - 4 = \pm 1$, which is $x = \pm\sqrt5$ and $x = \pm\sqrt3$.

Reciprocal trigonometric functions

The three reciprocal functions are defined wherever the trigonometric function in the denominator is not zero.

Here $n$ is any integer. Each asymptote sits where the denominator is zero: $\cos x = 0$ at $x = \dfrac{\pi}{2} + n\pi$, and $\sin x = 0$ at $x = n\pi$.

Since $|\sin x| \le 1$ and $|\cos x| \le 1$, the reciprocals satisfy $|\sec x| \ge 1$ and $|\operatorname{cosec} x| \ge 1$, so these graphs never enter the strip $-1 < y < 1$. The graph of $y = \operatorname{cosec} x$ touches the graph of $y = \sin x$ at every point where $\sin x = \pm 1$, that is at $x = \dfrac{\pi}{2} + n\pi$. There the cosecant curve has a local minimum at height $1$ or a local maximum at height $-1$, sitting in the U-shaped pieces between the asymptotes. Likewise $y = \sec x$ touches $y = \cos x$ where $\cos x = \pm 1$, at $x = n\pi$.

The graph of $y = \cot x$ is decreasing on each interval between asymptotes and crosses the $x$-axis where $\cos x = 0$, at $x = \dfrac{\pi}{2} + n\pi$. It is the reciprocal of $\tan x$, so its zeroes are the asymptotes of $\tan x$ and its asymptotes are the zeroes of $\tan x$.

The reciprocal functions have their own Pythagorean identities, found by dividing $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$ or by $\sin^2 x$:

The graph of y = |f(x)|

The absolute value of $f(x)$ is $f(x)$ where $f(x) \ge 0$ and $-f(x)$ where $f(x) < 0$. On the graph, every part that lies below the $x$-axis is reflected in the $x$-axis, and every part on or above the axis stays where it is. The result lies entirely on or above the $x$-axis.

For $f(x) = x^2 - 4$, the parabola dips below the axis for $-2 < x < 2$, with a minimum at $(0, -4)$. The graph of $y = |x^2 - 4|$ flips that part up, so it has a local maximum at $(0, 4)$ and sharp corners at $(\pm 2, 0)$.

The graph of y = f(|x|)

Here $x$ is replaced by its absolute value, so the function takes the same value at $x$ and at $-x$ and the graph is symmetric about the $y$-axis. The part of the graph of $y = f(x)$ with $x \ge 0$ is kept, the part with $x < 0$ is discarded, and the kept part is reflected in the $y$-axis to replace it.

For $f(x) = x - 1$, the graph of $y = |x| - 1$ is a V shape with its vertex at $(0, -1)$, crossing the $x$-axis at $x = \pm 1$. This differs from $y = |x - 1|$, which has its vertex at $(1, 0)$, so it matters which bars are where.

Adding and subtracting functions

The sum $f(x) + g(x)$ and difference $f(x) - g(x)$ are defined only where both $f$ and $g$ are defined, so the domain of the new function is the intersection of the two domains. Given $f(x) = x + 1$ and $g(x) = x - 3$ with domain all real $x$, the difference is $f(x) - g(x) = (x + 1) - (x - 3) = 4$, which has a horizontal graph at height $4$.

When the functions are given only as graphs, each ordinate of the sum is found by adding the two heights at the same value of $x$, and only over the interval where both graphs exist. If $f$ is defined for $x \ge 0$ and $g$ for $x \le 5$, the sum is defined for $0 \le x \le 5$. Where one graph is zero, the sum follows the other, and where the two heights are equal and opposite, the sum is zero.