Expert: Paul Klarreich - 1/18/2008

Question
Hi,
If I have a function y=(3x-5)^k, where k is +ve and k>0
When will I have inflection points based on whether k is odd or even

thanks!

Answer
Questioner:   vince
Category:  Calculus
Private:  No
 
Subject:  calculus - second derivative, inflection point
Question:  Hi,
If I have a function y=(3x-5)^k, where k is +ve and k>0
These say the same thing: --------------^^^^^^ ---- ^^^


When will I have inflection points based on whether k is odd or even

thanks!

.........................................
Hi, Vince,

You can easily send questions here. It is rare for me to be maxed out, because I am listed in Calculus and Advanced Math.  If I'm booked up in one, try the other.

......................................
An inflection point, by definition, separates an interval where  f'' > 0 from an interval where   f'' < 0.

Since  f'' = 9k(k-1)(3x - 5)^k-2, the only possible I.P. is at x = 5/3.

Now if x < 5/3, then 3x < 5, and 3x-5 < 0. (neg)
And if x > 5/3, then 3x > 5, and 3x-5 > 0. (pos)

And:           
Now if x < 5/3, f'' =  9k(k-1)(neg)^k-2
And if x > 5/3, f'' =  9k(k-1)(pos)^k-2

Does f'' change sign at x = 5/3.  
YES, if  k-2 is odd, which means k is odd.
NO, if  k-2 is even, which means k is even.

That's your conclusion.  x = 5/3 is an I.P, iff k is odd.

Extra note: This works for k >= 3.  If k = 1 or k = 2, there is no I.P.
('iff' is read: 'if and only if')