Kenneth S. answered • 11/28/17

Calculus will seem easy if you have the right tutor!

^{n -2}

Amber P.

asked • 11/28/17I also have to explain my reasoning

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Kenneth S. answered • 11/28/17

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Calculus will seem easy if you have the right tutor!

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.

Chris M. answered • 11/28/17

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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)=0^{n}=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)=x^{n} when n is an odd number!

Good Luck.

Cheers

-Chris

Michael J. answered • 11/28/17

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Kenneth S.

11/28/17