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Math Help - Need Help finding critical points and Inflection points. (graphing)

  1. #1
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    Need Help finding critical points and Inflection points. (graphing)

    The question asks for the critical points and inflection points of:

    (x^2-1)/(x^3+1)

    I've tried to take the derivitave and second derivitave but it just got too complicated. If anyone could help me that would be great.

    My quick commands menu doesn't seem to be working but the question is quite simple.

    Thanks!
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  2. #2
    Super Member Quacky's Avatar
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    Quote Originally Posted by mattyc33 View Post
    The question asks for the critical points and inflection points of:

    (x^2-1)/(x^3+1)

    I've tried to take the derivitave and second derivitave but it just got too complicated. If anyone could help me that would be great.

    My quick commands menu doesn't seem to be working but the question is quite simple.

    Thanks!
    I don't see any shorthand, I think we're just going to have to use the quotient rule.

    So Let u=x^2-1

    \frac{du}{dx}=2x

    Let v=x^3+1

    \frac{dv}{dx}=3x^2

    We have:

    \dfrac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

    =\dfrac{(x^3+1)(2x)-(x^2-1)(3x^2)}{(x^3+1)^2}

    =\dfrac{2x^4+2x-3x^4+3x^2}{x^6+2x^3+1}

    =\dfrac{-x^4+3x^2+2x}{x^6+2x^3+1}

    =\dfrac{x(-x^3+3x+2)}{(x^3+1)^2}

    \dfrac{-x(x-2)(x+1)(x+1)}{((x+1)(x^2-x+1))^2}

    =\dfrac{-x(x-2)}{(x^2-x+1)^2}

    ...which I suppose you could differentiate again, although there's got to be a faster approach? If there is, it's at a level that's beyond me, though.
    Last edited by Quacky; June 9th 2011 at 12:20 PM. Reason: Sign error, whoops!
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  3. #3
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    expression can written as

    y = x^5 - x^3 + x^2 - 1

    then y' = 5x^4 - 3x^2 + 2x

    for critical points 5x^4 - 3x^2 + 2x = 0 gives two real roots x = 0 and x = -1 now you can do it
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  4. #4
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    Quote Originally Posted by waqarhaider View Post
    expression can written as

    y = x^5 - x^3 + x^2 - 1

    then y' = 5x^4 - 3x^2 + 2x

    for critical points 5x^4 - 3x^2 + 2x = 0 gives two real roots x = 0 and x = -1 now you can do it
    Unfortunatly, I have no idea how you got that expression .
    When I plugged what Quacky said into my calculator I got two x values of x=0 and x=2.

    Could anyone please explain how he got this expression .
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  5. #5
    Super Member Quacky's Avatar
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    Quote Originally Posted by mattyc33 View Post
    Unfortunatly, I have no idea how you got that expression .
    When I plugged what Quacky said into my calculator I got two x values of x=0 and x=2.

    Could anyone please explain how he got this expression .
    Editted out a sign error with my original response - and to be honest I'm not entirely sure how he got there either.
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  6. #6
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    Thanks alot for clearing up the critical points.

    I'm pretty terrible at my derivitave calculation though so once again i'm having alot of trouble finding the second derivitave in order to find the inflection points.

    As Quacky pointed out the first derivitave is -x(x-2)/(x^2-x+1)^2

    If anyone could help me that would be great .
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  7. #7
    Super Member Quacky's Avatar
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    Quote Originally Posted by mattyc33 View Post
    Thanks alot for clearing up the critical points.

    I'm pretty terrible at my derivitave calculation though so once again i'm having alot of trouble finding the second derivitave in order to find the inflection points.

    As Quacky pointed out the first derivitave is -x(x-2)/(x^2-x+1)^2

    If anyone could help me that would be great .
    Again, I think there must be a shorter method, but what have you tried? Let u=-x^2+2x and v=(x^2-x+1)^2
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  8. #8
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    Yeah, I've tried using the quotient rule, it just gets very messy and I'm 99% sure i've been making mistakes in this mess 0.0
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  9. #9
    Super Member Quacky's Avatar
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    Quote Originally Posted by mattyc33 View Post
    Yeah, I've tried using the quotient rule, it just gets very messy and I'm 99% sure i've been making mistakes in this mess 0.0
    Please post some effort. Doing the first one took me long enough!
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  10. #10
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    Well all I have to say is that there must be an easier way:

    (-x^2+2x)'(x^2-x+1)^2 - (x^2+2x)(x^2-x+1)' / (x^2-x+1)^4

    (-2x+2)(x^2-x+1)^2 - (x^2+2x)[2(x^2-x+1) x (2x-1)] / (x^2-x+1)^4

    (-2x+2)(x^4-2x^3+3x^2-2x+1) - (x^2+2x)(4x-2)(x^2-x+1) / (x^2-x+1)^4

    (-2x^5+2x^4-6x^3+4x^2-2x+2x^4-4x^3+6x^2-4x+2) - (4x^5-4x^3-4x) / (x^2-x+1)^4

    And from there... I have no idea .

    I'm not saying to try it haha, I know there must be another way, thanks for the help!
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  11. #11
    Super Member Quacky's Avatar
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    Indeed: The roots are horrid!
    http://www.wolframalpha.com/input/?i=Differentiate+-x%28x-2%29%2F%28x^2-x%2B1%29^2
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  12. #12
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    Quote Originally Posted by Quacky View Post
    Indeed: The roots are horrid!
    http://www.wolframalpha.com/input/?i=Differentiate+-x%28x-2%29%2F%28x^2-x%2B1%29^2
    Whew thats great thanks!
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  13. #13
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    Try this:

    \frac{x^{2}-1}{x^{3}+1}=\frac{(x-1)(x+1)}{(x+1)(x^{2}-x+1)}=\frac{x-1}{x^{2}-x+1},

    for x\not=-1.

    I think that might simplify things for you a bit.
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  14. #14
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    Sorry for the first reply where I missed '/' sign and solved (x^2-1) (x^3+1)
    Here is new reply:
    expression can be written as

    y = ( x - 1 ) / ( x^2 - x + 1 )

    then y' = [ ( x^2 - x + 1 ) 1 - ( x - 1 ) ( 2x - 1 ) ] / ( x^2 - x + 1 )^2

    gives y' = ( 2x - x^2 ) / ( x^2 - x + 1 )

    again y" = [ ( x^2 - x + 1 )^2 ( 2 - 2x ) - ( 2x - x^2 ) 2( x^2 - x + 1 )( 2x - 1 ) ]/ ( x^2 -x + 1 )^4

    gives y" = ( 2x^3 - 6x^2 + 2 ) / ( x^2 - x + 1 )^3 and we find point of inflection
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  15. #15
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    Your derivatives look correct to me. So what do you get for critical points and points of inflection?
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