Find the points of inflection fx 9x x2 4 ItSolution fx 9x

Find the points of inflection.

f(x) = __9x__
x^2 + 4

It

f(x)= _9x__ x^2 + 4 f\'(x)=(9(x^2 +4)-9x(2x))/(x^2 +4)^2=9(4-x^2)/(x^2 +4)^2 f\'\'(x)=9 (-2x(x^2+4)^2 -2x(4-x^2)(x^2 +4))/(x^2 +4)^4 =0 ==>9(x^2 +4)(-2x^3-8x-8x+2x^3)=0 ==>x=0 ==>f(x)=0 so (0,0) is inflection point

$Find the points of inflection. f(x) = __9x__ x^2 + 4 ItSolution f(x)= _9x__ x^2 + 4 f\'(x)=(9(x^2 +4)-9x(2x))/(x^2 +4)^2=9(4-x^2)/(x^2 +4)^2 f\'\'(x)=9 (-2x(x^2$