Point of inflection

[mathpat]

Homework Statement

x^2 / x-1. Identify any asymptotes, extrema and points of inflection.

Homework Equations

The Attempt at a Solution

I am stuck trying to derive my first derivative. My first derivative equals x(x-2)/(x-1)^2. I tried to use the quotient rule again using while incorporating the chain rule and after multiple attempts I kept resulting with 2 / (x-1)^3.

Any suggestions?

 

[STEMucator]

This should help in general, switch your quotient rule into a product & chain rule :

$\frac{x^2}{x-1} = x^2 (x-1)^{-1}$

 

[Mark44]

Homework Statement

x^2 / x-1. Identify any asymptotes, extrema and points of inflection.

Write this as an equation, and use parentheses.

f(x) = x²/(x - 1)

Homework Equations

The Attempt at a Solution

I am stuck trying to derive my first derivative. My first derivative equals x(x-2)/(x-1)^2.

As an equation, this is f'(x) = x(x-2)/(x-1)².

I tried to use the quotient rule again using while incorporating the chain rule and after multiple attempts I kept resulting with 2 / (x-1)^3.

Any suggestions?

Your answer is correct, and this can be verified by calculating the derivative using the technique that Zondrina suggested.

 

[mathpat]

I see... so basically that means there is no point of inflection since I can not set that equation equal to zero and solve?


By the way, I appreciate both of your help.

 

[Mark44]

I see... so basically that means there is no point of inflection since I can not set that equation equal to zero and solve?

The sign of f'' changes at x = 1. If x < 1, f''(x) < 0, and if x > 1, f''(x) > 0. Does this mean that there is an inflection point at x = 1? Why or why not?

By the way, I appreciate both of your help.