Question

$f(x)=sin^{-1}(sec^{-1}x)$

Answer 1

[1, sec(1)]

Answer 2

The given function is $f(x) = \sin^{-1}(\sec^{-1}x)$.

For the function $f(x)$ to be defined, two conditions must be satisfied:

1. The argument of the outer function, $\sin^{-1}$, must be within its domain. The domain of $\sin^{-1}(u)$ is $[-1, 1]$. Thus, we must have $\sec^{-1}x \in [-1, 1]$.
2. The argument of the inner function, $\sec^{-1}$, must be within its domain. The domain of $\sec^{-1}(x)$ is $(-\infty, -1] \cup [1, \infty)$. Thus, we must have $x \in (-\infty, -1] \cup [1, \infty)$.

Let's analyze the first condition: $\sec^{-1}x \in [-1, 1]$. The principal range of the function $\sec^{-1}x$ is $[0, \pi] \setminus \{\frac{\pi}{2}\}$. We need the values of $x$ such that $\sec^{-1}x$ falls within the interval $[-1, 1]$. Let $y = \sec^{-1}x$. We require $y \in [-1, 1]$. Also, from the definition of $\sec^{-1}x$, we know that $y$ must be in the range $[0, \pi] \setminus \{\frac{\pi}{2}\}$. So, we need $y \in [-1, 1] \cap ([0, \pi] \setminus \{\frac{\pi}{2}\})$. The intersection of the interval $[-1, 1]$ and the set $[0, \pi] \setminus \{\frac{\pi}{2}\}$ is $[0, 1]$. This is because the range $[0, \pi] \setminus \{\frac{\pi}{2}\}$ contains only non-negative values (from 0 up to $\pi \approx 3.14$, excluding $\pi/2 \approx 1.57$). The interval $[-1, 1]$ contains values from -1 to 1. The common interval is $[0, 1]$. So, the condition $\sec^{-1}x \in [-1, 1]$ is equivalent to $\sec^{-1}x \in [0, 1]$. We need to find the values of $x$ such that $0 \le \sec^{-1}x \le 1$.

Let $y = \sec^{-1}x$. We have $0 \le y \le 1$. By the definition of the inverse secant function, $x = \sec y$, and $y \in [0, \pi] \setminus \{\frac{\pi}{2}\}$. The interval $[0, 1]$ is a subset of $[0, \pi/2)$ since $1$ radian $\approx 57.3^\circ$ and $\pi/2$ radians $\approx 90^\circ$. In the interval $[0, \pi/2)$, the function $\sec y$ is strictly increasing. Therefore, if we apply the secant function to the inequality $0 \le y \le 1$, the inequality signs are preserved:

$\sec(0) \le \sec(y) \le \sec(1)$

$1 \le x \le \sec(1)$

Now we must consider the second condition for the domain of $f(x)$: $x$ must be in the domain of $\sec^{-1}x$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$.

We need to find the values of $x$ that satisfy both conditions:

1. $1 \le x \le \sec(1)$
2. $x \in (-\infty, -1] \cup [1, \infty)$

We need to find the intersection of the interval $[1, \sec(1)]$ and the set $(-\infty, -1] \cup [1, \infty)$. Since $1$ radian is in the first quadrant ($0 < 1 < \pi/2$), $\cos(1)$ is positive and $0 < \cos(1) < 1$. $\sec(1) = \frac{1}{\cos(1)}$. Since $0 < \cos(1) < 1$, we have $\sec(1) > 1$. So, the interval $[1, \sec(1)]$ starts at 1 and extends to a value greater than 1. This interval $[1, \sec(1)]$ is completely contained within the set $[1, \infty)$. The intersection of $[1, \sec(1)]$ and $(-\infty, -1] \cup [1, \infty)$ is $[1, \sec(1)]$.

Thus, the domain of the function $f(x) = \sin^{-1}(\sec^{-1}x)$ is $[1, \sec(1)]$.