Answer
$$\eqalign{
& {\text{Inflection points:}} \cr
& \left( { - 2.87, - 0.293} \right),\left( {0.5320,0.844} \right),\left( { - 0.6527,0.4491} \right) \cr
& {\text{Vertical asymptotes: none}} \cr
& {\text{Horizontal asymptote at }}y = 0 \cr
& {\text{Relative maximum at }}\left( {0,1} \right) \cr
& {\text{Relative minimum at }}\left( { - 2, - \frac{1}{3}} \right) \cr
& {\text{Domain}}:\left( { - \infty ,\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{x + 1}}{{{x^2} + x + 1}} \cr
& {\text{First set the domain}}{\text{,}} \cr
& {x^2} + x + 1 \ne 0 \cr
& {\text{Where }}{x^2} + x + 1{\text{ is}} \ne 0{\text{ for all real numbers}}{\text{, }} \cr
& {\text{then the domain of the function is: }}D:\left( { - \infty ,\infty } \right) \cr
& \cr
& {\text{*Differentiating}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{x + 1}}{{{x^2} + x + 1}}} \right] \cr
& f'\left( x \right) = \frac{{\left( {{x^2} + x + 1} \right)\left( 1 \right) - \left( {x + 1} \right)\left( {2x + 1} \right)}}{{{{\left( {{x^2} + x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{{x^2} + x + 1 - 2{x^2} - x - 2x - 1}}{{{{\left( {{x^2} + x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{ - {x^2} - 2x}}{{{{\left( {{x^2} + x + 1} \right)}^2}}} \cr
& {\text{Set }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr
& \frac{{ - {x^2} - 2x}}{{{{\left( {{x^2} + x + 1} \right)}^2}}} = 0 \cr
& - {x^2} - 2x = 0 \cr
& x\left( {x + 2} \right) = 0 \cr
& {x_1} = 0,\,{\text{ }}{x_2} = - 2 \cr
& \cr
& {\text{*Find the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - {x^2} - 2x}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}} \right] \cr
& {\text{Using a CAS we obtain}} \cr
& f''\left( x \right) = \frac{{2\left( {{x^3} + 3{x^2} - 1} \right)}}{{{{\left( {{x^2} + x + 1} \right)}^3}}} \cr
& \cr
& {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}{x_1} = 0,\,{\text{ }}{x_2} = - 2 \cr
& {\text{Using the second derivative test}}{\text{, we conclude that}} \cr
& f''\left( 0 \right) = \frac{{2\left( {0 + {0^2} - 1} \right)}}{{{{\left( {{0^2} + 0 + 1} \right)}^3}}} = - 2 < 0{\text{ Relative maximum }} \cr
& f''\left( { - 2} \right) = \frac{{2\left( {{{\left( { - 2} \right)}^3} + 3{{\left( { - 2} \right)}^2} - 1} \right)}}{{{{\left( {{{\left( { - 2} \right)}^2} + \left( { - 2} \right) + 1} \right)}^3}}} = \frac{{38}}{{343}} > {\text{Relative minimum }} \cr
& f\left( 0 \right) = 1 \to {\text{Relative maximum at }}\left( {0,1} \right) \cr
& f\left( { - 2} \right) = - \frac{1}{3} \to {\text{Relative minimum at }}\left( { - 2, - \frac{1}{3}} \right) \cr
& \cr
& {\text{Set }}f''\left( x \right) = 0 \cr
& \frac{{2\left( {{x^3} + 3{x^2} - 1} \right)}}{{{{\left( {{x^2} + x + 1} \right)}^3}}} = 0 \cr
& {x^3} + 3{x^2} - 1 = 0 \cr
& {\text{Solving by using technology }} \cr
& {x_1} = - 2.87,{\text{ }}{x_2} = 0.5320,{\text{ }}{x_3} = - 0.6527 \cr
& {\text{The inflection points are:}} \cr
& \left( { - 2.87, - 0.293} \right),\left( {0.5320,0.844} \right),\left( { - 0.6527,0.4491} \right) \cr
& \cr
& {\text{There are no vertical asymptotes because the denominator}} \cr
& {\text{is never 0}}{\text{.}} \cr
& \cr
& *{\text{Find the horizontal asymptotes}} \cr
& \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{x + 1}}{{{x^2} + x + 1}} = 0 \cr
& \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{x + 1}}{{{x^2} + x + 1}} = 0 \cr
& {\text{Horizontal asymptote at }}y = 0 \cr
& \cr
& {\text{Summary}} \cr
& {\text{Inflection points:}} \cr
& \left( { - 2.87, - 0.293} \right),\left( {0.5320,0.844} \right),\left( { - 0.6527,0.4491} \right) \cr
& {\text{Vertical asymptotes: none}} \cr
& {\text{Horizontal asymptote at }}y = 0 \cr
& {\text{Relative maximum at }}\left( {0,1} \right) \cr
& {\text{Relative minimum at }}\left( { - 2, - \frac{1}{3}} \right) \cr
& {\text{Domain}}:\left( { - \infty ,\infty } \right) \cr
& \cr
& {\text{Graph}} \cr} $$
