pt of inflexion (1 Viewer)

[mojako]

if f(x) has a pt of inflexion at x = a,
will the curve of "1 / f(x)" have a pt of inflexion at x = a??
Thanks.

 

[mervvyn]

no, not necessarily. points of inflexion occur when the second derivative = 0, and f(x) and 1/f(x) will have different derivatives which potentially have different zeroes.

 

[Xayma]

Not necessairly. Consider the case of f(x)=x³
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)]²
g''(x)={2[f'(x)]²f(x)-[f(x)]²f''(x)}/[f(x)]⁴

So if f(a)≠0 it will if 2[f'(a)]²=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.

 

[mojako]

no, not necessarily. points of inflexion occur when the second derivative = 0, and f(x) and 1/f(x) will have different derivatives which potentially have different zeroes.

very true.. how could I not think about that...

To Xayma: thanks for dedicating ur time to answer my question