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Mathematics · Ch 4 — Inverse Trigonometric Functions

Graph of the Inverse Secant Function

4.7.3

Graph of the Inverse Secant Function

y=sec⁡−1xy=\sec^{-1}x has domain R∖(−1,1)\mathbb{R}\setminus(-1,1) and range [0,π]∖{π2}[0,\pi]\setminus\left\{\tfrac{\pi}2\right\} — that is, sec⁡−1:R∖(−1,1)→[0,π]∖{π2}\sec^{-1}:\mathbb{R}\setminus(-1,1)\to[0,\pi]\setminus\left\{\tfrac{\pi}2\right\}.

Fig. 4.25 shows the restricted secant curve on its principal domain, and Fig. 4.26 its reflection in y=xy=x: two flattening branches, one for x≥1x\ge1 approaching the horizontal asymptote y=π2y=\tfrac{\pi}2 and starting at (1,0)(1,0), one for x≤−1x\le-1 approaching y=π2y=\tfrac{\pi}2 from the other side and reaching up to y=πy=\pi, with a gap in the graph directly above (−1,1)(-1,1). …

Figure 4.25Graph of y = sec x on the restricted domain [0, pi] excluding pi/2: a branch over [0, pi/2) rising from 1 to infinity and a branch over (pi/2, pi] rising from minus infinity to -1; vertical asymptote at x = pi/2.
Fig. 4.25 — Graph of y = sec x on the restricted domain [0, pi] excluding pi/2: a branch over [0, pi/2) rising from 1 to infinity and a branch over (pi/2, pi] rising from minus infinity to -1; vertical asymptote at x = pi/2.

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. Two branches meeting the horizontal lines y=1y=1 and y=−1y=-1 at x=0x=0 and x=πx=\pi respectively, each shooting to ±∞\pm\infty as x→π/2x\to\pi/2. …

Figure 4.26Graph of y = sec^{-1} x on domain R minus (-1,1): right branch from (1, 0) increasing toward pi/2, left branch from (-1, pi) decreasing toward pi/2; horizontal asymptote y = pi/2, range [0, pi] minus {pi/2}.
Fig. 4.26 — Graph of y = sec^{-1} x on domain R minus (-1,1): right branch from (1, 0) increasing toward pi/2, left branch from (-1, pi) decreasing toward pi/2; horizontal asymptote y = pi/2, range [0, pi] minus {pi/2}.

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. Two flattening curve branches for x≥1x\ge1 and x≤−1x\le-1 approaching the horizontal asymptotes y=π/2y=\pi/2 and y=πy=\pi (with y=0y=0 at x=1x=1), and a gap directly above the interval (−1,1)(-1,1). …