Question

Find the critical points of this function.

Answer 1

I understand the calculus, but I just forgot the algebra haha.
\[f(x) = \frac{ \sqrt{\ln (x)} }{ x }\]

Which I derive into \[\frac{ \frac{ 1 }{ 2\sqrt{\ln(x)}} -\sqrt{\ln(x)} }{ x^{2} } = 0\] and set f'(x) as 0.

Then I completely forgot how to deal with the natural logarithm. -_-

Answer 2

Ok so now let's multiply both sides of that thing by \(x^2\) which will clean it up a bit:
\[ \frac{ 1 }{ 2\sqrt{\ln(x)}} -\sqrt{\ln(x)} = 0\]

I'd say make this temporary substitution so you're not distracted by it, then plug it back in to deal with it later:

\[a = \sqrt{\ln x}\]

Now solve for a:

\[\frac{1}{2a} - a = 0\]

Answer 3

\[\sqrt{1/2}\]?

Answer 4

@Kainui

Answer 5

Yeah, looks good, keep going!

Answer 6

Coming up soon, the inverse function of \(\ln x\) is \(e^x\) so:
\[e^{\ln x} = x\]
It also works the other way around, in case you were curious.
\[\ln(e^x) = x\]

Answer 7

Do exponents cancel each other out?

Answer 8

I think they do right? If they're equal?

Answer 9

Can I do
\[(\ln x)^\frac{ 1 }{ 2 } = \frac{ 1 }{ 2 } ^ \frac{ 1 }{ 2 }\]
So that\[x = e^\frac{ 1 }{ 2 }\]

Answer 10

Ah yes in this way they do yep. But don't think of it as them cancelling each other out really, it's more like if you square both sides of this equation, since both sides are square roots, you get the regular thing back.

So like when x is positive, \((\sqrt{x})^2 = x\)

Answer 11

Oh yeah that makes sense.

Answer 12

So \[e^{\frac{ 1 }{ 2 }}\]
is correct?

Answer 13

Yeah I believe so, you could plug it in to check I guess if you doubt it. I am kinda distracted but I think that's what I got. I smashed a huge spider that was in my bed on the paper I worked this out with a little bit ago so my work is gone.

Answer 14

lol ok. Beautiful. Thanks!