A necessary condition for the function to have an inflection point is that the second derivative vanishes.
Calculate the second derivative:
#f(x) = frac (x+4) ((x-2)^2)#
#f'(x) = frac ((x-2)^2-2(x-2)(x+4)) ((x-2)^4) = frac ((x-2)-2(x+4)) ((x-2)^3) = frac (x-2-2x-8) ((x-2)^3)= -frac (x+10) ((x-2)^3)#
#f''(x) = frac (-(x-2)^3+3(x+10)(x-2)^2) ((x-2)^6) = frac (-(x-2)+3(x+10)) ((x-2)^4) = frac (-x+2+3x+30) ((x-2)^4) = 2 frac (x+16) ((x-2)^4)#
So #f''(x) = 0# for #x=-16#.
To be sure this is effectively an inflection point we have to check that #f''(x)# changes sign around #x=-16#. We can see that the denominator is always positive, and the numerator is of first grade in x, so it definitely changes sign around the root.
We can conclude that #x=-16# is an inflection point.
