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

1. ## 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.

2. $f'(x)=\ln{x}+1$

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

$\mbox{Critical Value}=0$

However, $\ln{x}>0$

$f''(1) >0$

Thus, the function is concave up.

3. Originally Posted by k101
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 $x\rightarrow\ \infty$

4. Thank you!

5. So this mean's that the critical value of f'(x) is 0? Also, how come we are evaluating at 1 for f"?

6. Originally Posted by k101
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.

7. How do we know it has no asymptotes? Could you explain please

8. Originally Posted by k101
How do we know it has no asymptotes? Could you explain please
Post 3.

9. 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.

10. Originally Posted by Archie Meade

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.

11. Ah! Ok, I was thinking of "critical value" only in terms of a turning point.

12. Originally Posted by k101
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).

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

14. Originally Posted by Pranas
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 ....?

15. Originally Posted by mr fantastic
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.