Mathematics - Functions Question with Solution | TestHub

We have, $f(x)=\frac{x^2+x+2}{x^2+x+1}=\frac{\left(x^2+x+1\right)+1}{x^2+x+1}=1+\frac{1}{{\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}}$

We can see here that as $\mathrm{x}\to \infty ,\mathrm{f}(\mathrm{x})\to 1$ which is the min value of $\mathrm{f}(\mathrm{x})$. Also $\mathrm{f}(\mathrm{x})$ is max when ${\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}$ is min which is so when $x=-\frac{1}{2}$ and then $\frac{3}{4}$. $\begin{array}{llc}\therefore & f_{\max }=1+\frac{1}{3/4}=\frac{7}{3}\therefore & R_j=\left(1,\frac{7}{3}\right]\end{array}$