Domain of One Variable Function – A Function with sin inside square root – Exercise 2446

Exercise

Determine the domain of the function:

f(x)=sin⁡xf(x)=\sqrt{\sin\sqrt{x}}

Final Answer


4π2k2≤x≤π2(1+2k)2,k∈N4\pi^2 k^2 \leq x\leq \pi^2(1+2k)^2, k\in N

Solution

Let’s find the domain of the function:

f(x)=sin⁡xf(x)=\sqrt{\sin\sqrt{x}}

There are square roots, so we need the expressions inside the roots to be non-negative:

x≤0  and  sin⁡x≤0x\leq 0\text{  and  }\sin \sqrt{x}\leq 0

Let’s look at the second inequality. The sin function is positive in the first half of its cycle, meaning when the following holds:

0≤x≤π0\leq \sqrt{x}\leq \pi

But sin function has endless cycles. Let’s express this with parameter k (integer). Adding the cycle in both sides will result in

2πk≤x≤π+2πk2\pi k \leq \sqrt{x}\leq \pi+2\pi k

2πk≤x≤π(1+2k)2\pi k \leq \sqrt{x}\leq \pi (1+2k)

4π2k2≤x≤π2(1+2k)24\pi^2 k^2 \leq x\leq \pi^2(1+2k)^2

Note that since x is between two quadratic expressions, x is necessarily positive or zero. That is, the first inequality also exists (had it not happened, we would have to intersect them). Therefore, that’s the final answer.

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