Not necessairly. Consider the case of f(x)=x<sup>3</sup>
It has a point of inflexion at x=0
However, 1/f(x) is not defined at x=0. Therefore it does not have a point of inflexion.
Otherwise if g(x)=1/f(x)
g'(x)=-f'(x)/[f(x)]<sup>2</sup>
g''(x)={2[f'(x)]<sup>2</sup>f(x)-[f(x)]<sup>2</sup>f''(x)}/[f(x)]<sup>4</sup>
So if f(a)≠0 it will if 2[f'(a)]<sup>2</sup>=f(a)f''(a)
Since f''(a)=0
f'(a)=0
∴ only if it is a horizontal point of inflexion.
So basically: If a curve f(x) has a point of inflexion at x=a, then the curve 1/f(x) will have a point of inflexion at x=a iff f(a)≠0 and it is a horizontal point of inflexion at x=a.
Hmm give me a sec Ill just check if there are any extra conditions using the third derivative method. Nope the above is true (at least for polynomials up to degree 5) since f'''(a)≠0 (up to degree 5) and f(a)≠0.