Let $f:ℝ\to ℝ$  be given by $f\left(x\right)=\left\{\begin{array}{cc}{x}^5+5x^4+10x^3+10x^2+3x+1,& x<0;\\ x^2-x+1,& 0\le x<1;\\ \frac{2}{3}{x}^3-4x^2+7x-\frac{8}{3},& 1\le x<3;\\ \left(x-2\right){\log }_e\left(x-2\right)-x+\frac{10}{3},& x\ge 3.\end{array}\right.$ Then which of the following options is/are correct?

Let $f:R\to R$

$\begin{array}{c}f\left(x\right)=\left\{\begin{array}{l}\begin{array}{c}{x}^5+5x^4+10x^3+10x^2+3x+1 ,x<0\\ x^2-x+1 ,0\le x<1 \end{array}\\ \begin{array}{c} \frac{2}{3}{x}^3-4x^2+7x-\frac{8}{3} ,1\le x<3\\ \left(x-2\right)\ln \left(x-2\right)-x+\frac{10}{3} ,x\ge 3\end{array}\end{array}\right.\\ =\left\{\begin{array}{cc}{\left(x+1\right)}^5-2x,& x<0\\ x^2-x+1,& 0\le x\le 1\\ \frac{2}{3}{x}^3-4x^2+7x-\frac{8}{3},& 1\le x<3\\ \left(x-2\right)\ln \left(x-2\right)-x+\frac{10}{3},& x\ge 3\end{array}\right.\\ f^1\left(x\right)=\left\{\begin{array}{cc}5{\left(x+1\right)}^4-2,& x<0\\ 2x-1,& 0\le x<1\\ 2x^2-8x+7,& 1\le x<3\\ 1+\ln \left(x-2\right)-1,& x⩾3\end{array}\right.\\ f^{11}\left(x\right)=\left\{\begin{array}{cc}20{\left(x+1\right)}^3,& x<0\\ 2& 0\le x<1\\ 4x-8& 1\le x<3\\ \frac{1}{x-2}& x⩾3\end{array}\right.\end{array}$

$if x\to -\infty then f\left(x\right)\to -\infty and x\to \infty \Rightarrow f\left(x\right)\to \infty$

The range is = R and f is onto

$\begin{array}{l}\text{ Clearly }{f}^1\left(-\frac{1}{2}\right)=\frac{5}{16}-2<0 \therefore f is not increa\sin g in \left(-\infty ,0\right)\\ \text{ Also }{f}^{11}(1-)=2>0\\ f^{11}(1+)=-4<0\mid \\ \therefore f^1\left(x\right)\text{ has local maximum at }x=1\\ \left.\begin{array}{c}{\left(f^1\right)}^1(1+)=-4\\ \left(f^1\right)(1-)=2\end{array}\right\}L.D\ne R,D\therefore f^1\text{ is not differentiable at }\mathrm{x}=1\end{array}$