SOLUTION : $f(x)=\frac{1}{2\{-x\}}+1-\{x\}-1=\frac{1}{2\{-x\}}+\{-x\}-1$
$\geq \sqrt{2}-1 \quad \text { (Using A.M. } \geq \text { G.M.) }$
Range of $f(x)$ is $[\sqrt{2}-1, \infty)$
$f(x)=\frac{\sqrt{x^2+1}-3 x}{\sqrt{x^2+1}+x}=\frac{1-\frac{3 x}{\sqrt{x^2+1}}}{1+\frac{x}{\sqrt{x^2+1}}}$
Let $\quad g(x)=\frac{x}{\sqrt{x^2+1}}$ and $h(x)=\frac{1-3 x}{1+x}$. Thus $f(x)=h(g(x))$
Now, $g^{\prime}(x)=\frac{\sqrt{x^2+1}-x \frac{1(2 x)}{2 \sqrt{x^2+1}}}{x^2+1}=\frac{1}{\left(x^2+1\right)^{3 / 2}}>0$ and $h^{\prime}(x)=\frac{(1+x)(-3)-(1-3 x)(1)}{(1+x)^2}=\frac{-4}{(1+x)^2}<0$ and $\quad f^{\prime}(x)=h^{\prime}(g(x)) g^{\prime}(x)<0$
$\Rightarrow \quad \operatorname{minimum} f(x)=\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty} \frac{\sqrt{x^2+1}-3 x}{\sqrt{x^2+1}+x}=\lim _{x \rightarrow \infty} \frac{\sqrt{1+\frac{1}{x^2}}-3}{\sqrt{1+\frac{1}{x^2}}+1}=-1 \\$
$\Rightarrow \quad \operatorname{maximum~} h(x)=\lim _{x \rightarrow-\infty} f(x)=\lim _{x \rightarrow-\infty} \frac{\sqrt{x^2+1}-3 x}{\sqrt{x^2+1}+x}=\lim _{x \rightarrow-\infty} \frac{|x| \sqrt{1+\frac{1}{x^2}}-3 x}{|x| \sqrt{1+\frac{1}{x^2}}+x} \\$
$=\lim _{x \rightarrow-\infty} \frac{-\sqrt{1+\frac{1}{x^2}}-3}{-\sqrt{1+\frac{1}{x^2}}+1}=\infty \quad \Rightarrow \quad \text { Range of } f(x)=(-1, \infty)$
