Concavity of a function

xxlvh · #1 $Old$ November 6th, 2009, 10:59 PM

I am to determine the inflection point and intervals where the graph is concave up and concave down, and I have gotten a wrong answer somehow. I'm guessing it's probably when I solved for y'', it was a very long derivative to simplify:

$y = \frac{ln(x)}{6\sqrt{x}}$

$y' = \frac{6\sqrt(x)-ln(x)3\sqrt{x}}{36x^2}$

$y'' = \frac{3\sqrt(x)ln(x)-12\sqrt{x}}{36x^3}$

$y''=0$ when $x=e^4$

VonNemo19 · #2 $Old$ November 6th, 2009, 11:11 PM

Quote:

Originally Posted by xxlvh $View Post$

I am to determine the inflection point and intervals where the graph is concave up and concave down, and I have gotten a wrong answer somehow. I'm guessing it's probably when I solved for y'', it was a very long derivative to simplify:

$y = \frac{ln(x)}{6\sqrt{x}}$

$y' = \frac{6\sqrt(x)-ln(x)3\sqrt{x}}{36x^2}$

$y'' = \frac{3\sqrt(x)ln(x)-12\sqrt{x}}{36x^3}$

$y''=0$ when $x=e^4$

$y'=\frac{2-\ln{x}}{12\sqrt[3]{x^2}}$

Debsta · #3 $Old$ November 6th, 2009, 11:16 PM

...which is equivalent to what xxlvh has above. This gives solution to y'=0 as x=e^2.
In von Nemo19's from it is much easier to find second deriv.

xxlvh · #4 $Old$ November 8th, 2009, 11:44 PM

I was able to find the inflection point but still having some difficulties with the concavity.

The simplified second derivative is $y'' = \frac{18\sqrt{x}ln(x)-48\sqrt{x}}{144x^3}$

Here's the inequalities I determined:
For it to be concave up,
$18\sqrt{x}ln(x)-48\sqrt{x}> 0$ and $144x^3 > 0$
or $18\sqrt{x}ln(x)-48\sqrt{x} < 0$ and $144x^3 < 0$

When I solved these I had $x < 0, x > e^{8/3}$

For it to be concave down,
$18\sqrt{x}ln(x)-48\sqrt{x}> 0$ and $144x^3 < 0$
or $18\sqrt{x}ln(x)-48\sqrt{x} < 0$ and $144x^3 > 0$

And the solution for this one didn't work out at all.