Results 1 to 15 of 15

Math Help - How do we find the intercepts and asymptotes for f(x) = x ln x?

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    12

    Exclamation How do we find the intercepts and asymptotes for f(x) = x ln x?

    How do we find the intercepts and asymptotes for f(x) = x ln x?

    Also, I am confused on why this function is concave up for all of the domain. Can someone please explain this, all help is greatly appreciated.
    Last edited by mr fantastic; January 14th 2011 at 03:35 AM. Reason: Coppied title into main body of post.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    f'(x)=\ln{x}+1

    \displaystyle f''(x)=\frac{1}{x}

    \mbox{Critical Value}=0

    However, \ln{x}>0

    f''(1) >0

    Thus, the function is concave up.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by k101 View Post
    Also, I am confused on why this function is concave up for all of the domain. Can someone please explain this, all help is greatly appreciated.
    Using the product rule

    f'(x)=\frac{x}{x}+lnx=1+lnx

    f''(x)=\frac{1}{x}>0 for x>0

    which means the tangent slope keeps increasing.

    x-intercept:

    xlnx=0\Rightarrow\ lnx=0\Rightarrow\ x=1

    Asymptote:

    The curve approaches no limit as x\rightarrow\ \infty
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2008
    Posts
    12
    Thank you!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2008
    Posts
    12
    So this mean's that the critical value of f'(x) is 0? Also, how come we are evaluating at 1 for f"?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by k101 View Post
    So this mean's that the critical value of f'(x) is 0? Also, how come we are evaluating at 1 for f"?
    When you do derivative tests, you evaluate numbers to the left and right of the critical value.

    Due to the domain restriction of the natural log, evaluating the left side of 0 isn't required.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2008
    Posts
    12
    How do we know it has no asymptotes? Could you explain please
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by k101 View Post
    How do we know it has no asymptotes? Could you explain please
    Post 3.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Critical value is unimportant.

    f'(x)=0\Rightarrow\ lnx=-1\Rightarrow\ x=\frac{1}{e}

    Due to the fact that the second derivative is positive, the curve is concave upward.

    The graph has no asymptotes.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Archie Meade View Post

    Due to the fact that the second derivative is positive, the curve is concave upward.
    That is what testing a point to right of zero (critical value) tells someone, and since there are no more points to test around greater than 0, it is true for all x > 0.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Ah! Ok, I was thinking of "critical value" only in terms of a turning point.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by k101 View Post
    How do we find the intercepts and asymptotes for f(x) = x ln x?

    Also, I am confused on why this function is concave up for all of the domain. Can someone please explain this, all help is greatly appreciated.
    Since \displaystyle \lim_{x \to 0} (x \ln(x)) = 0 the function has a 'hole' (removable discontinuity) at (0, 0).

    As has already been remarked, the x-intercept is at (1, 0) since the solution to 0 = x \ln(x) is x = 1.

    As has already been remarked, from calculus a minimum turning is found to be at (1/e, -1/e).

    The graph is concave up for x > 0 since f''(x) > 0 for x > 0.

    There are no asymptotes since the function is defined for all x > 0 (therefore no vertical asymptote) and \displaystyle \lim_{x \to +\infty} (x \ln(x)) = +\infty (therefore no horizontal asymptote).
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member Pranas's Avatar
    Joined
    Oct 2010
    From
    Europe. Lithuania.
    Posts
    81
    As for horizontal, non-vertical, oblique, slope asymptotes I believe we should be checking
    \[\mathop {\lim }\limits_{x \to  + \infty } (\frac{{f(x)}}{x}) = \mathop {\lim }\limits_{x \to  + \infty } (\frac{{x\ln (x)}}{x}) = \mathop {\lim }\limits_{x \to  + \infty } (\ln (x)) =  + \infty \]
    Last edited by Pranas; January 14th 2011 at 02:03 PM. Reason: Possible clarification
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Pranas View Post
    As for horizontal asymptotes I believe we should be checking
    \[\mathop {\lim }\limits_{x \to + \infty } (\frac{{f(x)}}{x}) = \mathop {\lim }\limits_{x \to + \infty } (\frac{{x\ln (x)}}{x}) = \mathop {\lim }\limits_{x \to + \infty } (\ln (x)) = + \infty \]
    So, from the above, f(x) = x has a horizontal asymptote ....?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member Pranas's Avatar
    Joined
    Oct 2010
    From
    Europe. Lithuania.
    Posts
    81
    Quote Originally Posted by mr fantastic View Post
    So, from the above, f(x) = x has a horizontal asymptote ....?
    I do not speak English miraculously well, sorry for that.
    My last comment was more likely meant to involve non-vertical asymptotes (oblique, slope asymptotes) and, since my learning source defines horizontal asymptote being just a case of oblique asymptotes, I came up with that.

    \displaystyle \[\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{f(x)}}{x}} \right) = 0\]
    would be the case when slope asymptote becomes parallel to X-axis.

    I didn't say somebody was wrong, but I don't see oblique asymptotes taken into account unless I'm missing something.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. intercepts and asymptotes
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: May 26th 2010, 09:36 PM
  2. Find the asymptotes and intercepts
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 19th 2009, 11:30 PM
  3. Replies: 1
    Last Post: October 25th 2008, 02:29 PM
  4. Replies: 3
    Last Post: July 13th 2008, 04:41 AM
  5. intercepts and asymptotes
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 17th 2008, 07:01 PM

Search Tags


/mathhelpforum @mathhelpforum