To identify inverse trig graphs, look at the intercepts: \((1, 0)\) is the signature of \(\sec^{-1} x\).
Check the domain: If the graph is missing between \(-1\) and \(1\), it must be \(\sec^{-1}\) or \(\text{cosec}^{-1}\).
Horizontal asymptotes indicate the bounds of the range.
Concept: • The domain of \(\sec^{-1} x\) is \((-\infty, -1] \cup [1, \infty)\).
• The principal value range of \(\sec^{-1} x\) is \([0, \pi] \setminus \{\frac{\pi}{2}\}\), but in some modified conventions, it is defined as \([-\pi, -\frac{\pi}{2}) \cup [0, \frac{\pi}{2})\).
• At \(x = 1\), \(y = \sec^{-1}(1) = 0\).
• At \(x = -1\), \(y = \sec^{-1}(-1) = \pi\) (Standard) or \(-\pi\) (Modified).
Step 1: Analyze the domain shown in the graph
The graph consists of two branches.
One branch exists for \(x \ge 1\) starting from the point \((1, 0)\).
The other branch exists for \(x \le -1\) starting from the point \((-1, 0)\).
This confirms the domain is \((-\infty, -1] \cup [1, \infty)\). This eliminates \(\tan^{-1} x\) (which has domain \(\mathbb{R}\)) and \(\cos^{-1} x\) (which has domain \([-1, 1]\)).
Step 2: Evaluate specific points to distinguish between \(\sec^{-1} x\) and \(\text{cosec}^{-1} x\)
The graph clearly passes through the point \((1, 0)\).
We check the values for \(\sec^{-1} x\) and \(\text{cosec}^{-1} x\):
For \(\sec^{-1} x\): \(\sec^{-1}(1) = 0\) because \(\sec(0) = 1\).
For \(\text{cosec}^{-1} x\): \(\text{cosec}^{-1}(1) = \frac{\pi}{2}\) because \(\text{cosec}(\frac{\pi}{2}) = 1\).
Since the graph includes \((1, 0)\), it represents a form of the \(\sec^{-1} x\) function.
Step 3: Examine the asymptotic behavior
The graph shows a horizontal asymptote at \(y = \frac{\pi}{2}\) as \(x \to \infty\).
It shows another horizontal asymptote at \(y = -\frac{\pi}{2}\) as \(x \to -\infty\).
This corresponds to the definition of \(\sec^{-1} x\) often used in calculus where the range is split to make the function's derivative always positive.
Thus, the graph represents \(y = \sec^{-1} x\).
