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  1. Jun 6, 2006 #1

    F.B

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    Im having a problem with curve sketching i cant figure out how to do this problem. I am confused.

    Heres the question

    1. For the following curve find:
    a)the domain
    b)the intercepts
    c)the asymptotes
    d)intervals of increase or decrease: local maximum and minimum
    e)concavity and the point of inflection

    x^2 - x - 1 / x-1


    Heres my work.

    a) the domain is x = 1
    b)Let y=0
    y = 0 - 0 - 1/ 0-1
    y=1
    Let x = 0
    x^2 - x - 1 = 0
    x = 1.62 or x = 0.62

    c) Vertical asymptote:
    x = 1
    Horizontal asymptote
    x^2 - x - 1 / x-1
    y = x - 1/ x-1
    Therefore it has an slant asympote at y = x

    Heres where it gets fun

    d) y = x^2 - x - 1/x-1
    y'= x^2 - 2x + 2/ (x-1)^2
    let y'= 0
    x^2 - 2x - 2 = 0
    x = 1 +/- i

    Do i say that it doesnt have an maximum or minimum point because they are
    imaginary?

    e) y'' = -2/(x-1)^3
    Let y''=0
    -2 cannot equal 0

    So there are no points of inflection, but then how do i determine the
    concavity when my first derivative is based on imaginary numbers.

    I have never encountered one like this. I dont know how to determine the intervals of decrease or increase because of i, so can you please help me.
     
  2. jcsd
  3. Jun 7, 2006 #2

    Andrew Mason

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    How do you get that? The domain is the set of all x where
    f(x) = x^2 - x - 1 / x-1

    So x = 1 is NOT part of the domain of f(x).
    You must mean let x = 0 since you are trying to find the y intercept. The y intercept is a point (0,y)
    Again, this should be y = 0 to find the x intercept.
    These are the correct solutions to the quadratic equation.
    Correct
    You will have to explain why you did this. I don't follow this.
    You lost me there.

    AM
     
  4. Jun 7, 2006 #3

    HallsofIvy

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    Assuming that y= x^2- x- 1/(x-1), this is exactly backwards: the domain is all x except x= 1.

    As Andrew Mason pointed out, actually, you've let x= 0 to find the y-intercept.

    No, you let y= 0 so you get the equation x^2- x- 1/(x-1)= 0 which is the same as x^2- x= 1/(x-1) or (x-1)(x^2-x)= (x-1)^2(x)= 0. The x intercepts are x= 0 and x= 1.

    Ok.

    But the problem didn't say "slant asymptote"- it asked about a horizontal asymptote- and there is none. No, y= x is not a "slant asymptote" anyway. How did you go from y= x^2- x- 1/(x-1) to y= x- 1/(x-1)??
    x^2- x is NOT equal to x!


    ?? Are you saying that the derivative of x^2 is x^2 and the derivative of -x is -2x? Surely you know better than that! Oh, and the derivative of -1/(x-1)= -(x-1)^(-1) is NOT 2/(x-1)^2, although I admit that's harder. For all of those you can use the fact that the derivative of
    x^n is nx^(n-1).

    Go back and do the derivative correctly!

    Again, go back and do the first derivative correctly. You can't get the second derivative until you do that.

    For future reference, if you had a problem where the derivative was never 0, such as f(x)= x, then the derivative always has the same sign and so there is just one "interval of decrease or increase".
     
  5. Jun 8, 2006 #4

    F.B

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    To find the y-intercept you have to solve for y correct? Thats why you let x=0 to find the find what y is when equals 0. To find the x-intercept you have to let y be 0 and then you solve for x to determine what x is when y is 0. And Halls of Ivy maybe i should have made the question a bit clearer. The function is this:

    y=(x^2-x-1)/(x-1), does this help?

    For the horizontal asymptote we have to find one even it is a slant one.
    By doing long division i get the following:

    y = (x) - (1)/(x-1).

    My derivatives are all right because when i use the question rule i get that same thing.

    But i figured how to find the intervals of increase and decrease. Because the vertical asymptote x=1, i can use those as my intervals. So x<1 and x>1.

    But i still have one problem. If i use those as my intervals, when i do that chart with the test values, do i sub a number <1 and >1 into the original function or the first derivative.

    My other problem is determining the concavity. Because i dont have a point of inflection, how would i determine where the function is concave up and down.
     
  6. Jun 8, 2006 #5

    Hootenanny

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    That is correct.

    Halls of Ivy's point still remains valid, the function is defined everywhere except at x = 1. Therefore, the domain of the function is [itex] x \in\Re \; , \; x \neq 1[/itex]

    I'm afraid you derivative is incorrect. Try and do it again, perhaps it would be easier to write it as;

    [tex]f(x) = (x^{2} - x -1)(x-1)^{-1}[/tex]
     
    Last edited: Jun 8, 2006
  7. Jun 8, 2006 #6

    TD

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    The concavity is given by the sign of the second derivative. Watch out though: the fact that there are no zeroes doesn't mean the concavity is the same everywhere, remember your vertical asymptote!
     
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