Answer
$$\left( {0,1} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2{x^5} - 10{x^4} + 20{x^3} + x + 1 \cr
& {\text{Calculate the second derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^5} - 10{x^4} + 20{x^3} + x + 1} \right] \cr
& f'\left( x \right) = 10{x^4} - 40{x^3} + 60{x^2} + 1 \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {10{x^4} - 40{x^3} + 60{x^2} + 1} \right] \cr
& f''\left( x \right) = 40{x^3} - 120{x^2} + 120x \cr
& {\text{Set }}f''\left( x \right) = 0 \cr
& 40{x^3} - 120{x^2} + 120x = 0 \cr
& 40x\left( {{x^2} - 3x + 3} \right) = 0 \cr
& {\text{Zero - factor property}} \cr
& 40x = 0,{\text{ }}{x^2} - 3x + 3 = 0 \cr
& x = 0 \cr
& {\text{Use the quadratic formula to solve }}{x^2} - 3x + 3 = 0 \cr
& x = \frac{{ - \left( { - 3} \right) \pm \sqrt {{{\left( { - 3} \right)}^2} - 4\left( 1 \right)\left( 3 \right)} }}{2} = \frac{{3 \pm \sqrt { - 3} }}{2},{\text{ no real solutions}} \cr
& {\text{then, }}x = 0 \cr
& {\text{The inflection point is }}\left( {0,f\left( 0 \right)} \right) \cr
& f\left( 0 \right) = 2{\left( 0 \right)^5} - 10{\left( 0 \right)^4} + 20{\left( 0 \right)^3} + \left( 0 \right) + 1 = 1 \cr
& {\text{Inflection point }}\left( {0,1} \right) \cr} $$
