I can't figure this interval out.

Let $\displaystyle f(x)=6x+\frac{1}{x}$ find:

$\displaystyle f'(x)>0$ for $\displaystyle x \in$ _____?

$\displaystyle f'(x)<0$ for $\displaystyle x \in$ _____?

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

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

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

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

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.

Re: I can't figure this interval out.

perhaps because you used $\displaystyle (-\infty, -\sqrt{\frac{1}{6}}] \cup [\sqrt{\frac{1}{6}}, \infty)$instead of $\displaystyle (-\infty, -\sqrt{\frac{1}{6}}) \cup (\sqrt{\frac{1}{6}}, \infty) $ (notice the ] replaced with ) ).at $\displaystyle \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

Re: I can't figure this interval out.

Thanks a ton jakncoke, I would gladly buy you one given the chance.

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