Results 1 to 12 of 12

Math Help - Application of Limits and Derivatives to Rational Functions

  1. #1
    Junior Member
    Joined
    Apr 2011
    Posts
    68

    Application of Limits and Derivatives to Rational Functions

    Hey guys, please tell me if I'm correct in my calculations
    Given the function \frac{x^2}{x-1}

    a) Determine the domain and all the intercepts of the function. Does the function have any symmetry?


    This is what I did: Domain is all real #s except 1, because x-1=0, x=1.

    intercepts are/is: (0,0) because \frac{(0)^2}{(0)-1}=0

    it's symmetric because \frac{(-x)^2}{(-x)-1} = \frac{x^2}{-x-1}

    Is this right?

    b) Determine \lim_{x \to \infty} f(x) and \lim_{x \to -\infty} f(x). Determine all horizontal and vertical asymptotes of the function.

    I got that lim x-->inf it goes to inf and lim x-->-inf goes to -inf.
    Vertical asymptote is 1 as shown above.

    Can someone show me how you get the horizontal asymptote for this please?

    c) find all critical numbers.

    Quotient rule:
    \frac{(x-1)(2x)-(x^2)(1)}{(x-1)^2} = \frac{x^2-2x}{(x-1)^2 }

    x^2-2x=0, x=0,2 critical numbers are 0 and 2

    d) Determine the intervals on which the function is increasing and where it is decreasing. Find all maximum and minimum values of f.

    I use a chart.
    ______0__2
    x __|- | + | +
    x-2 |- | - | +
    f '_ |+ | - | +
    ___up, down, up. so there must be a maximum at 0 and a minimum and 2. right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1
    Quote Originally Posted by RezMan View Post
    Hey guys, please tell me if I'm correct in my calculations
    Given the function \frac{x^2}{x-1}

    a) Determine the domain and all the intercepts of the function. Does the function have any symmetry?


    This is what I did: Domain is all real #s except 1, because x-1=0, x=1.

    intercepts are/is: (0,0) because \frac{(0)^2}{(0)-1}=0

    it's symmetric because \frac{(-x)^2}{(-x)-1} = \frac{x^2}{-x-1}

    Is this right?
    A function is even if f(-x) = f(x) and odd if f(-x) = -f(x). In your case
    f(-x) = \frac{x^2}{-x - 1}

    Is this even, odd, or neither?

    Quote Originally Posted by RezMan View Post

    b) Determine \lim_{x \to \infty} f(x) and \lim_{x \to -\infty} f(x). Determine all horizontal and vertical asymptotes of the function.

    I got that lim x-->inf it goes to inf and lim x-->-inf goes to -inf.
    Vertical asymptote is 1 as shown above.

    Can someone show me how you get the horizontal asymptote for this please?
    [COLOR=Red][COLOR=Black]
    You get a horizontal asymptote when \lim_{x \to \pm \infty}f(x) \to \text{constant}. If this case your limits are either positive or negative infinity. So you have no horizontal asymptotes.

    -Dan

    Edit: You do have a "slant" asymptote if you need to look for one of those. The slant asymptote is y = x + 1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1
    Quote Originally Posted by RezMan View Post
    c) find all critical numbers.

    Quotient rule: [/COLOR][/COLOR] \frac{(x-1)(2x)-(x^2)(1)}{(x-1)^2} = \frac{x^2-2x}{(x-1)^2 }

    x^2-2x=0, x=0,2 critical numbers are 0 and 2
    Don't forget to set your denominator to 0. x = 1 is also a critical number.


    Quote Originally Posted by RezMan View Post
    d) Determine the intervals on which the function is increasing and where it is decreasing. Find all maximum and minimum values of f.

    I use a chart.
    ______0__2
    x __|- | + | +
    x-2 |- | - | +
    f '_ |+ | - | +
    ___up, down, up. so there must be a maximum at 0 and a minimum and 2. right?
    Redo your chart with the x = 1 in there.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2011
    Posts
    68
    Thank you so much Dan!
    Quote Originally Posted by topsquark View Post
    Is this even, odd, or neither?
    I guess it's neither, right? I just thought there was some rule where you only pay attention to the top.

    Quote Originally Posted by topsquark View Post
    You do have a "slant" asymptote if you need to look for one of those. The slant asymptote is y = x + 1.
    Can you please please show me how you got that?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1
    Quote Originally Posted by RezMan View Post
    Thank you so much Dan! I guess it's neither, right? I just thought there was some rule where you only pay attention to the top.
    That's correct, it is neither.

    Quote Originally Posted by RezMan View Post
    Can you please please show me how you got that?
    A slant asymptote, otherwise known as an "oblique" asymptote, is similar to a horizontal asymptote, except that it is off at an angle. (I'll leave the pictures to the link.)

    To sum up, you can tell if a function has a slant asymptote by looking at the degree of the numerator (dn) and the degree of the denominator (dd). If dn - dd = 1 then there is a slant asymptote. We get the form of the asymptote by division.

    For example, in your function you have
    f(x) = \frac{x^2}{x - 1}

    The degree of the numerator is 2 and the denominator 1. 2 - 1 = 1 so you have a slant asymptote. Unfortunately I can't code the long division, so I'll have to leave that to you. Here's the result.
    f(x) = \frac{x^2}{x - 1} = x + 1 + \frac{1}{x - 1}

    As with the test for a horizontal asymptote we look at the form as x goes to infinity. The 1/(x - 1) term becomes inconsequential, so the result is y = x + 1. That's your slant asymptote.

    -Dan

    Edit: You can actually do the division without using long division. Notice that
    \frac{x^2}{x - 1} = \frac{x^2 - 1 + 1}{x - 1} = \frac{x^2 - 1}{x - 1} + \frac{1}{x - 1} = (x + 1) + \frac{1}{x - 1}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Apr 2011
    Posts
    68
    Quote Originally Posted by topsquark View Post
    Redo your chart with the x = 1 in there.
    ______0__1__2
    x __|- | + | + |+
    x-2 |- | - | - | +
    f '_ |+ | - | - | +
    __up, down, down, up. so there's local maximum at 0 and local minimum at 2, and no absolute max/min?

    I can't thank you enough, this exercise is so difficult for me. There's more

    e) Determine where f is concave up and concave down.

    I figured that f''' =  \frac{2}{(x-1)^3} . so when plugging in 0 f'' is concave down, and concave up at 2.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1
    Quote Originally Posted by RezMan View Post
    ______0__1__2
    x __|- | + | + |+
    x-2 |- | - | - | +
    f '_ |+ | - | - | +
    __up, down, down, up. so there's local maximum at 0 and local minimum at 2, and no absolute max/min?

    I can't thank you enough, this exercise is so difficult for me. There's more

    e) Determine where f is concave up and concave down.

    I figured that f''' =  \frac{2}{(x-1)^3} . so when plugging in 0 f'' is concave down, and concave up at 2.
    The chart looks good now. A point: A function is concave up or down on an interval. So f(x) is concave up whenever f''(x) is negative and concave down when f''(x) is positive.

    You can either figure that out by punching numbers into your second derivative (or sketch it, the second derivative is easy enough to sketch) or by using your chart. You know where the turning points are and you know what the slope is doing. So, for example, the slope is increasing on (-infinity, 0), it has a local maximum at x = 0 and then decreases on (0, 1). What must the concavity of the function be on (-infinity , 1)? Think about this for the second branch of the function, (1, infinity).

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Apr 2011
    Posts
    68
    is the minima/maxima thing I said correct though? That there's a local min&max but no absolute?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1
    Quote Originally Posted by RezMan View Post
    is the minima/maxima thing I said correct though? That there's a local min&max but no absolute?
    When f''(x) is positive, the function is concave down. When f''(x) is negative, the function is concave up. It is the reverse of what you might expect. So you have your concavities reversed.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Apr 2011
    Posts
    68
    Quote Originally Posted by topsquark View Post
    When f''(x) is positive, the function is concave down. When f''(x) is negative, the function is concave up. It is the reverse of what you might expect. So you have your concavities reversed.
    -Dan
    Are you sure? Because the graph makes sense and also this website says the opposite
    Concavity of Graphs
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1
    Quote Originally Posted by RezMan View Post
    Are you sure? Because the graph makes sense and also this website says the opposite
    Concavity of Graphs
    (sighs) I think what I was reaching for is that we have a local minimum on a concave up domain and a local maximum on a concave down domain. Sorry for the poor information!

    -Dan
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Apr 2011
    Posts
    68
    Don't hit yourself against the wall dan!

    So are you saying that my min/max is wrong?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivatives of Rational Functions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 9th 2010, 06:50 PM
  2. Replies: 1
    Last Post: February 16th 2010, 08:00 AM
  3. Replies: 4
    Last Post: January 8th 2010, 03:30 AM
  4. Rational Functions Application Question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 29th 2009, 11:02 PM
  5. Replies: 3
    Last Post: March 12th 2009, 02:20 PM

/mathhelpforum @mathhelpforum