# I can't figure this interval out.

• Feb 20th 2013, 09:06 PM
KhanDisciple
I can't figure this interval out.
Let $f(x)=6x+\frac{1}{x}$ find:
$f'(x)>0$ for $x \in$ _____?
$f'(x)<0$ for $x \in$ _____?

for the derivative I got $6-\frac{1}{x^2}$.

I set that to zero to find the critical points, and found $\pm \sqrt{\frac{1}{6}}$

For $f'(x)>0$ I got $(-\infty,-\sqrt{\frac{1}{6}}\ ] \cup \[ \sqrt{\frac{1}{6}},\infty )$
For $f'(x)<0$ I got $[ -\sqrt{\frac{1}{6}},0 ) \cup (0,\sqrt{\frac{1}{6}} ]$

I was told that these are incorrect, and I don't understand why. I even graphed the derivative and this looked to me to be correct. Am I close? Please let me know where I am wrong. I really appreciate it. Thanks.
• Feb 20th 2013, 09:42 PM
jakncoke
Re: I can't figure this interval out.
perhaps because you used $(-\infty, -\sqrt{\frac{1}{6}}] \cup [\sqrt{\frac{1}{6}}, \infty)$instead of $(-\infty, -\sqrt{\frac{1}{6}}) \cup (\sqrt{\frac{1}{6}}, \infty)$ (notice the ] replaced with ) ).at $\pm \sqrt{\frac{1}{6}}$ f'(x) is 0, it shouldnt be included as a value for which f'(x) > 0.

Same reasoning for your interval for f'(x) < 0
• Feb 20th 2013, 10:00 PM
KhanDisciple
Re: I can't figure this interval out.
Thanks a ton jakncoke, I would gladly buy you one given the chance.
• Feb 21st 2013, 10:16 PM
EliteAndoy
Re: I can't figure this interval out.
I think you missed one critical point, that is where f'(x) is undefined. which is when x=0. So if you evaluate the first derivative test. The function is decreasing at (0,sqrt(1/6)) and increasing at (sqrt(1/6),infinity) because remember that any value of x less than or equal to zero is now allowed to be used as a critical point. Try this one out and hope this helps! :D