What will be the domain and range of $f(x)=\sqrt{\sin^{-1}({\frac{1}{1+3x+2x^{2}}})}$?

Question

If $f(x)=\sqrt{\sin^{-1}({\frac{1}{1+3x+2x^{2}}})}$ , then what will be the domain and range of $f(x)$ ? I am trying this question by finding the domain of $f(x)$ first.

So, $2x^{2}+3x+1\neq0$ . Now $2x^{2}+3x+1=(2x+1)(x+1)$ . Therefore $x\neq-1/2,x\neq-1$ . Now lastly what I can guess is that $-1\leq\frac{1}{2x^{2}+3x+1}\leq1$ . But I can't proceed further.

Please help me out with this question.

Answer 1

First off, $x\longmapsto \sqrt{x}$ is only defined for $x\geq 0$.

For the domain of $x\longmapsto\sqrt{\arcsin x}$, this means that we need $\arcsin{x}\geq 0$, which is equivalent to $x\geq 0$.

So for the domain of $f(x):=\sqrt{\arcsin \left(\frac{1}{2x^2+3x+1}\right)}$, this means that $\frac{1}{2x^2+3x+1}\geq 0$. Also, the function $y\longmapsto \arcsin y$ is only defined for $y\in[-1,1]$, so we actually have two inequalities $$\frac{1}{2x^2+3x+1}\geq 0\wedge -1\leq \frac{1}{2x^2+3x+1}\leq 1$$ which can be combined into $$0 \leq \frac{1}{2x^2+3x+1}\leq 1.$$

This is equivalent to $2x^2+3x+1\geq 1$, which can be reformulated as $$2x^2+3x\geq 0 \Leftrightarrow x(2x+3)\geq 0 \Leftrightarrow x\geq 0 \vee x\leq -\frac{3}{2}.$$

Thus, the domain of definition is $D:=(-\infty,-3/2]\cup [0,\infty)$.

For the range, notice that $x\longmapsto \sqrt{x}$ and $x\longmapsto\arcsin x$ are strictly increasing functions, whereas the inner-most function $x\longmapsto \frac{1}{2x^2+3x+1}$ is strictly increasing for $x\in(-\infty,-3/2]$ and strictly decreasing for $x\in[0,\infty)$. Thus, the composition $f$ of the three functions attains its maximal value either at $-3/2$ or at $0$. Calculation shows that both values are actually the same, namely $\sqrt{\frac{\pi}{2}}$.

Further, $\lim\limits_{x\rightarrow\pm\infty} \frac{1}{2x^2+3x+1}=0$, so $\lim\limits_{x\rightarrow\pm\infty}\arcsin\left( \frac{1}{2x^2+3x+1}\right)=0$ and also $\lim\limits_{x\rightarrow\pm\infty} f(x)=0$. Due to continuity, every value between $0$ and the maximal value is attained, so the range is $\left(0,\sqrt{\frac{\pi}{2}}\right]$.

Answer 2

HINT.-The domain and rank of $\sin^{-1}(x)$ are $[-1,1]$ and $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ respectively so, because of the square root, we need that $$0\le \sin^{-1}\left(\frac{1}{(2x+1)(x+1)}\right)\le\frac{\pi}{2}$$ which is equivalent to $$0\le \left(\frac{1}{(2x+1)(x+1)}\right)\le1$$ It follows that the required domain is $D=[-\infty,-1.5]\cap[0,\infty]$ and the rank can be calculate with $\sqrt{\sin^{-1}(1)}\approx1.2533124$ so the rank is equal to $[0,1.2533124]$.