Answer
$$\eqalign{
& {\text{max}} = - 2 - 2\sqrt 2 {\text{ at }}x = - 1 - \sqrt 2 . \cr
& {\text{There is no minimum value}}{\text{.}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^2} + 1}}{{x + 1}};{\text{ }}\left( { - 5,1} \right) \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{{\left( {x + 1} \right)\left( {2x} \right) - \left( {{x^2} + 1} \right)\left( 1 \right)}}{{{{\left( {x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{2{x^2} + 2x - {x^2} - 1}}{{{{\left( {x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{{x^2} + 2x - 1}}{{{{\left( {x + 1} \right)}^2}}} \cr
& f'\left( x \right) = 0 \cr
& \frac{{{x^2} + 2x - 1}}{{{{\left( {x + 1} \right)}^2}}} = 0 \cr
& {x^2} + 2x - 1 = 0 \cr
& {\text{By the quadratic formula}} \cr
& x = \frac{{ - 2 \pm \sqrt {{{\left( 2 \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{2} \cr
& x = \frac{{ - 2 \pm \sqrt 8 }}{2} = \frac{{ - 2 \pm 2\sqrt 2 }}{2} \cr
& x = - 1 \pm \sqrt 2 \cr
& {\text{Both values are on the interval }}\left( { - 5,1} \right) \cr
& {\text{Evaluate }}f{\text{ at all the critical points and at the end points }}a{\text{ and }}b \cr
& f\left( { - 5} \right) = - \frac{{13}}{2} = - 6.5 \cr
& f\left( { - 1 - \sqrt 2 } \right) = - 2 - 2\sqrt 2 \approx - 4.828 \cr
& f\left( { - 1 + \sqrt 2 } \right) = - 2 + 2\sqrt 2 \approx 0.82842 \cr
& f\left( 1 \right) = 1 \cr
& {\text{The largest value of }}f{\text{ in the interval }}\left( { - 5,1} \right){\text{ is }} - 2 + 2\sqrt 2 ,{\text{ so}} \cr
& {\text{there is a max}} = - 2 - 2\sqrt 2 {\text{ at }}x = - 1 - \sqrt 2 . \cr
& {\text{There is no minimum value}}{\text{.}} \cr} $$