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.

- January 13th 2011, 04:32 PMk101How 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. - January 13th 2011, 04:35 PMdwsmith

However,

Thus, the function is concave up. - January 13th 2011, 04:37 PMArchie Meade
- January 13th 2011, 04:38 PMk101
Thank you!

- January 13th 2011, 04:40 PMk101
So this mean's that the critical value of f'(x) is 0? Also, how come we are evaluating at 1 for f"?

- January 13th 2011, 04:41 PMdwsmith
- January 13th 2011, 04:47 PMk101
How do we know it has no asymptotes? Could you explain please

- January 13th 2011, 04:49 PMdwsmith
- January 13th 2011, 04:50 PMArchie Meade
Critical value is unimportant.

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

The graph has no asymptotes. - January 13th 2011, 04:53 PMdwsmith
- January 13th 2011, 05:02 PMArchie Meade
Ah! Ok, I was thinking of "critical value" only in terms of a turning point.

- January 14th 2011, 03:32 AMmr fantastic
Since 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 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 (therefore no horizontal asymptote). - January 14th 2011, 05:42 AMPranas
As for horizontal, non-vertical, oblique, slope asymptotes I believe we should be checking

- January 14th 2011, 01:37 PMmr fantastic
- January 14th 2011, 02:03 PMPranas
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.

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.