Thus, the function is concave up.
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.
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).
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.