To find the list of POSSIBLE inflection points we need to first compute the second derivative of f(x) with respect to x.
Using the power rule:
The possible inflection points are found where f''(x) is either undefined or where f''(x)=0.
Since f''(x) is a polynomial it is continuous and hence its defined for all x. So lets find the points where f''(x)=0
so set f''(x)=0=n(n-1)xn-2
dividing through by n(n-1) we have xn-2=0 which means x=0. This is our only POSSIBLE inflection point.
Note that f(0)=0n=0 so our only possible inflection point is indeed the origin (0,0).
Now, in order for x=0 to be an inflection point we need to assess the concavity on either side of x=0. If the concavity changes from one side of x=0 to the other than we have an inflection point.
Lets pick simple values on either side of x=0, say x=-1 and x=1
Note that f''(1)=(1)n-2=1>0 for ALL values of the integer n, meaning f(x) is concave up when x>0.
If x=0 is going to be an inflection point than f''(x)<0 (ie concave down) when x<0
We'll test this at f''(-1)=(-1)n-2
So now finally we have to find the values of integer n where (-1)n-2<0
Lets test some values
for n=0, (-1)n-2=(-1)0-2=(-1)-2=1/(-1)2=1>0 so that doesn't work
for n=1, (-1)n-2=(-1)1-2=-1<0 so that DOES work
for n=2, (-1)n-2=(-1)2-2=1>0 nope
If you test n=3,4,... and also n=-1,-2,...you can verify that f''(x)<0 when n is an odd number!
Hence (0,0) is an inflection point for f(x)=xn when n is an odd number!