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.

- Jan 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. - Jan 13th 2011, 04:35 PMdwsmith
$\displaystyle f'(x)=\ln{x}+1$

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

$\displaystyle \mbox{Critical Value}=0$

However, $\displaystyle \ln{x}>0$

$\displaystyle f''(1) >0$

Thus, the function is concave up. - Jan 13th 2011, 04:37 PMArchie Meade
Using the product rule

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

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

which means the tangent slope keeps increasing.

x-intercept:

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

Asymptote:

The curve approaches no limit as $\displaystyle x\rightarrow\ \infty$ - Jan 13th 2011, 04:38 PMk101
Thank you!

- Jan 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"?

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

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

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

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

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

- Jan 14th 2011, 03:32 AMmr fantastic
Since $\displaystyle \displaystyle \lim_{x \to 0} (x \ln(x)) = 0$ 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 $\displaystyle 0 = x \ln(x)$ 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 $\displaystyle \displaystyle \lim_{x \to +\infty} (x \ln(x)) = +\infty$ (therefore no horizontal asymptote). - Jan 14th 2011, 05:42 AMPranas
As for horizontal, non-vertical, oblique, slope asymptotes I believe we should be checking

$\displaystyle \[\mathop {\lim }\limits_{x \to + \infty } (\frac{{f(x)}}{x}) = \mathop {\lim }\limits_{x \to + \infty } (\frac{{x\ln (x)}}{x}) = \mathop {\lim }\limits_{x \to + \infty } (\ln (x)) = + \infty \]$ - Jan 14th 2011, 01:37 PMmr fantastic
- Jan 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.

$\displaystyle \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.