Finding the Domain of a Function: Step-by-Step Analysis

To find the domain of function \(f(x) = \sqrt {\cos (\sin x)} + {\sin ^{ - 1}}\left( {\frac{{1 + {x^2}}}{{2x}}} \right)\), we need to determine the values of \(x\) for which both parts of the function are defined. The function is a sum of two terms: \(g(x) = \sqrt {\cos (\sin x)}\) and \(h(x) = {\sin ^{ - 1}}\left( {\frac{{1 + {x^2}}}{{2x}}} \right)\). The domain of \(f(x)\) is the intersection of the domains of \(g(x)\) and \(h(x)\).

Domain of the First Term: \(g(x) = \sqrt {\cos (\sin x)}\)

For the square root to be defined, the expression inside the square root must be non-negative. So, we need \(\cos (\sin x) \ge 0\).

Let \(\theta = \sin x\). The range of \(\sin x\) for real \(x\) is \([-1, 1]\). So, we need to find the values of \(\theta \in [-1, 1]\) such that \(\cos \theta \ge 0\).

The cosine function \(\cos \theta\) is non-negative for \(\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\) plus multiples of \(2\pi\). Specifically, for \(\theta\) around 0, \(\cos \theta \ge 0\) when \(-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}\).

Let's compare the interval \([-1, 1]\) (the range of \(\sin x\)) with the interval \(\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\) (where cosine is non-negative). Note that \(\frac{\pi}{2} \approx 1.57\). So, the interval \([-1, 1]\) is completely contained within \(\left( - \frac{\pi}{2}, \frac{\pi}{2} \right)\). That is, \(-1 > -\frac{\pi}{2}\) and \(1 < \frac{\pi}{2}\).

Since for any real \(x\), \(\sin x \in [-1, 1]\), and for any \(\theta \in [-1, 1]\), \(\cos \theta\) is positive (specifically, \(\cos \theta > 0\) because \(\theta\) is strictly between \(-\pi/2\) and \(\pi/2\) except potentially at the endpoints, but \(\cos(-1)\) and \(\cos(1)\) are positive). Therefore, \(\cos(\sin x) > 0\) for all real \(x\). The condition \(\cos(\sin x) \ge 0\) is always satisfied for all \(x \in \mathbb{R}\).

Thus, the domain of \(g(x) = \sqrt {\cos (\sin x)}\) is all real numbers, \(x \in \mathbb{R}\).

Domain of the Second Term: \(h(x) = {\sin ^{ - 1}}\left( {\frac{{1 + {x^2}}}{{2x}}} \right)\)

For the inverse sine function, \(\sin^{-1}(u)\), to be defined, the argument \(u\) must satisfy \(-1 \le u \le 1\). Here, \(u = \frac{{1 + {x^2}}}{{2x}}\).

So, we need to solve the inequality: $$-1 \le \frac{{1 + {x^2}}}{{2x}} \le 1$$ Also, the expression \(\frac{{1 + {x^2}}}{{2x}}\) must be defined, which means \(2x \ne 0\), so \(x \ne 0\).

We can split the inequality into two parts:

1. \(\frac{{1 + {x^2}}}{{2x}} \le 1\)
2. \(\frac{{1 + {x^2}}}{{2x}} \ge -1\)

Let's solve the first inequality: \(\frac{{1 + {x^2}}}{{2x}} \le 1\)

The term \((x-1)^2\) is always non-negative (\(\ge 0\)). For the fraction to be less than or equal to zero, the denominator \(2x\) must be negative (since the numerator is \(\ge 0\)). However, the numerator is zero when \(x=1\), which makes the fraction \(0\). So, either \(2x < 0\) (which means \(x < 0\)) or \((x-1)^2 = 0\) (which means \(x=1\)).

If \(x < 0\), the inequality \(\frac{{(x - 1)^2}}{{2x}} \le 0\) holds because \((x-1)^2 > 0\) (since \(x \ne 1\)) and \(2x < 0\).

If \(x = 1\), \(\frac{{(1 - 1)^2}}{{2(1)}} = \frac{0}{2} = 0\), which satisfies \(0 \le 0\).

So, the solution to the first inequality \(\frac{{1 + {x^2}}}{{2x}} \le 1\) is \(x \in (-\infty, 0) \cup \{1\}\).

Now, let's solve the second inequality: \(\frac{{1 + {x^2}}}{{2x}} \ge -1\)

The term \((x+1)^2\) is always non-negative (\(\ge 0\)). For the fraction to be greater than or equal to zero, the denominator \(2x\) must be positive (since the numerator is \(\ge 0\)). However, the numerator is zero when \(x=-1\), which makes the fraction \(0\). So, either \(2x > 0\) (which means \(x > 0\)) or \((x+1)^2 = 0\) (which means \(x=-1\)).

If \(x > 0\), the inequality \(\frac{{(x + 1)^2}}{{2x}} \ge 0\) holds because \((x+1)^2 > 0\) (since \(x \ne -1\)) and \(2x > 0\).

If \(x = -1\), \(\frac{{(-1 + 1)^2}}{{2(-1)}} = \frac{0}{-2} = 0\), which satisfies \(0 \ge 0\).

So, the solution to the second inequality \(\frac{{1 + {x^2}}}{{2x}} \ge -1\) is \(x \in (0, \infty) \cup \{-1\}\).

The domain of \(h(x)\) is the intersection of the solutions to both inequalities, keeping in mind \(x \ne 0\):

Let's look at the intersections:

• The intersection of \((-\infty, 0)\) and \((0, \infty)\) is empty.
• The intersection of \((-\infty, 0)\) and \(\{-1\}\) is \(\{-1\}\).
• The intersection of \(\{1\}\) and \((0, \infty)\) is \(\{1\}\).
• The intersection of \(\{1\}\) and \(\{-1\}\) is empty.

So, the intersection is \(\{-1\} \cup \{1\} = \{-1, 1\}\). This is the domain for the second term involving \(\arcsin (1+x^2)/2x\).

Combining the Domains

The domain of function \(f(x)\) is the intersection of the domain of \(g(x)\) and the domain of \(h(x)\).

• Domain of \(g(x)\): \(x \in \mathbb{R}\)
• Domain of \(h(x)\): \(x \in \{-1, 1\}\)

Intersection: \(\mathbb{R} \cap \{-1, 1\} = \{-1, 1\}\).

Therefore, the function \(f(x)\) is defined for \(x \in \{-1, 1\}\).

This analysis confirms the domain of function f(x) is restricted to the set \(\{-1, 1\}\) based on the requirements for the \(\arcsin\) term, as the \(\sqrt{\cos(\sin x)}\) term is defined for all real numbers. Understanding how to find the domain of function is crucial in mathematics and calculus.