Point of inflection

  • Thread starter mathpat
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  • #1
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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?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
STEMucator
Homework Helper
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This should help in general, switch your quotient rule into a product & chain rule :

##\frac{x^2}{x-1} = x^2 (x-1)^{-1}##
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
35,431
7,296
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.
 

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