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.
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.
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 $\displaystyle x\rightarrow\ \infty$
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).
As for horizontal, non-vertical, oblique, slope asymptotes I believe we should be checking
$\displaystyle \[\mathop {\lim }\limits_{x \to + \infty } (\frac{{f(x)}}{x}) = \mathop {\lim }\limits_{x \to + \infty } (\frac{{x\ln (x)}}{x}) = \mathop {\lim }\limits_{x \to + \infty } (\ln (x)) = + \infty \]$
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.