Are There Inflection Points for the Curve 1/x - 1/(x-1)?

[Haroldoo]

Homework Statement

The question is: Find the Inflection Point(s) and Intervals of concavity for the curve

Homework Equations

Equation of Curve: 1/x - 1/(x-1)

The Attempt at a Solution

Ok so to find the Inflection point, we have been taught to find the second derivative of the equation, equate it to zero then solve for x. Ok so I tried this and I just can't seem to solve for X, the equation becomes enormous and unruly and I just don't know how to solve it. Anyways, I was looking at the graph and I couldn't see anywhere that would be an inflection point. So my question is, is there even any inflection points on this graph, and if the answer is no, then does that mean that there are no intervals of concavity as well?

Edit: Scratch what I said about no concavity, I think just by looking at the graph that it is Concave down when: x<0, x>1 and it it Concave up when: 0<x<1. However, my question about there being no inflection point still stands.
Thanks in advance =D

 

[GoldPheonix]

Well, it is a function of an inverse function (however more complicated), and since it is the summation of two inverse functions, they will retain the properties of the inverse functions, no?

Inverse functions have no Points of Inflection equal to 0. Depending on your definition, most people would define it as when the second derivative equals zero or infinity. Well, look at your second derivative:

y'' = 2/x^3 - 2/(x-1)^3

Is there anyway that these two equations can ever equal each other and cancel out to zero or infinity?

1/x^3 = 1/(x-1)^3

What's logical now? Flip them.

x^3 = (x - 1) ^3

This is some complex foiling, but if you recall your Newton's Binomial Theorem and Pascal's Triangle, it isn't too bad.

x^3 = x^3 - x^2 + x - 1

This should be obvious. =P

0 = -x^2 + x - 1

You'll need the Quadratic Formula:

0= -(x^2 - x + 1)

(1 +- sqrt(-3))/2 deals with imaginary numbers, therefore the root DNE.

So, there is no PoI for this function.

 

[Haroldoo]

Wow I pretty much love you right now haha, thanks a lot for your help. I don't really know what I was doing when I was trying to find the second derivative, your way was much much better. You explained it very well, you should be a teacher or somthing lol. Anyways, thanks again =D

 

[HallsofIvy]

Well, it is a function of an inverse function (however more complicated), and since it is the summation of two inverse functions, they will retain the properties of the inverse functions, no?

Inverse functions have no Points of Inflection equal to 0. Depending on your definition, most people would define it as when the second derivative equals zero or infinity. Well, look at your second derivative:

y'' = 2/x^3 - 2/(x-1)^3

Is there anyway that these two equations can ever equal each other and cancel out to zero or infinity?

1/x^3 = 1/(x-1)^3

What's logical now? Flip them.

x^3 = (x - 1) ^3

This is some complex foiling, but if you recall your Newton's Binomial Theorem and Pascal's Triangle, it isn't too bad.

x^3 = x^3 - x^2 + x - 1

This should be obvious. =P

I hope it's not obvious! Using Pascal's triangle, as you say,
(x-1)³= x³- 3x²+ 3x- 1.

0 = -x^2 + x - 1

You'll need the Quadratic Formula:

0= -(x^2 - x + 1)

(1 +- sqrt(-3))/2 deals with imaginary numbers, therefore the root DNE.

So, there is no PoI for this function.

From x³= (x-1)³= x³- 3x²+ 3x- 1 you get 3x²- 3x+ 1= 0.