Verified answer

I suppose you mean f(x)=x√(x²+2x+3)

The inflection points of the graph of a twice differentiable function are at the points where its second derivative vanishes. If y=x√(x²+2x+3), then

y' = x'√(x²+2x+3) + x(√(x²+2x+3))' = √(x²+2x+3) + x(2x+2)/2√(x²+2x+3) =

= [(x²+2x+3) + x(x+1)]/√(x²+2x+3) = (2x²+3x+3)/√(x²+2x+3)

y'' = [(4x+3)√(x²+2x+3) - (2x²+3x+3)·(2x+2)/(2√(x²+2x+3)]/√(x²+2x+3)² =

= [(4x+3)(x²+2x+3) - (2x²+3x+3)·(x+1)]/√(x²+2x+3)^(3/2) =

= [2x³+6x²+12x+6]/√(x²+2x+3)^(3/2)

The denominator is never 0, since x²+2x+3=(x+1)²+2, so the inflection points are where the numerator is 0. Dividing by 2, this gives x³+3x²+6x+3=0, or (x+1)³+3(x+1)=1, which has only one solution, viz. x=-0.6778..., for which y=0,4793.

[My arithmetic may be wrong, but that's the idea.]