Mathematics · Ch 4 — Inverse Trigonometric Functions

Graph of the Inverse Cotangent Function

4.8.3

Graph of the Inverse Cotangent Function

$y=\cot^{-1}x$ has domain $\mathbb{R}$ and range $(0,\pi)$ — that is, $\cot^{-1}:\mathbb{R}\to(0,\pi)$ , the only one of the six inverse trig functions defined for EVERY real number with no gap in either domain or output.

Fig. 4.29 shows the restricted cotangent curve on $(0,\pi)$ , and Fig. 4.30 its reflection in $y=x$ : a strictly DECREASING, flattening curve running from just below $\pi$ (as $x\to-\infty$ ) through $\left(0,\tfrac{\pi}2\right)$ down to just above $0$ (as $x\to\infty$ ), hugging the two horizontal asymptotes $y=0$ and $y=\pi$ . …

Figure 4.29Graph of y = cot x on the restricted domain (0, pi): a single strictly decreasing branch from infinity through (pi/2, 0) to minus infinity; vertical asymptotes at x = 0 and x = pi.

Fig. 4.29 — Graph of y = cot x on the restricted domain (0, pi): a single strictly decreasing branch from infinity through (pi/2, 0) to minus infinity; vertical asymptotes at x = 0 and x = pi.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. A single steeply falling branch from $+\infty$ near $x=0$ to $-\infty$ near $x=\pi$ , crossing $y=0$ at $x=\pi/2$ . …

Figure 4.30Graph of y = cot^{-1} x on domain (-infinity, infinity): a strictly decreasing S-curve from pi (as x tends to minus infinity) through (0, pi/2) to 0 (as x tends to infinity); range (0, pi).

$Fig. 4.30 — Graph of y = cot^{-1} x on domain (-infinity, infinity): a strictly decreasing S-curve from pi (as x tends to minus infinity) through (0, pi/2) to 0 (as x tends to infinity); range (0, pi).$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. A strictly decreasing flattening curve running from just below $\pi$ (as $x\to-\infty$ ) through $\left(0,\pi/2\right)$ down to just above $0$ (as $x\to\infty$ ), hugging horizontal asymptotes $y=0$ and $y=\pi$ . …

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