Consider the function f(x)=x^n for positive integer values of n. (b) For what values of n does the function have a point of inflection at the origin?

Question

I also have to explain my reasoning

Answer 1

if n=1, the graph is a st line and has no inflection

if n=2, the parabola has no inflection point (f" is never zero); concavity is upward everywhere

if n=3, the cubic has an inflection point at the origin.

 

f" = n(n-1)x^{n -2}

In general, f" = 0 if n=1 but there is no inflection point because slope is constant everywhere.

When n is > 1, the concavity is downward at the left of the origin and positive to the right of the origin and zero when x = 0 so then there's inflection there. But if n is even, inflection does not occur because f" will always be positive.

 

NOTE: Michael J's answer should exclude n = 1, but is correct for odd n> 2.

 

 

Answer 2

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:

 

f'(x)=nx^n-1

f''(x)=n(n-1)x^n-2

 

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)x^n-2

 

dividing through by n(n-1) we have x^n-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!

 

Good Luck.

 

Cheers

 

-Chris

Answer 3

A cubic polynomial has a point of inflection at x=0.  So does a 5th degree polynomial.  Therefore, all positive odd integers of n.