Question

need help finding critical points :)

Answer 1

Hi, I need to find the critical points for f(x)= \[\frac{ \ln(x) }{ x^6}\]

Answer 2

I did the first step, which is to find the derivative. I got f ' (x)= \[\frac{ x^5-6xln(x) }{ x^12 }\]

Answer 3

the 12 looks weird on the bottom.... I know that one of the critical points is 0, but I'm confused about finding the other one

Answer 4

after factoring out an x from the top, you get
\[\large x^4-6ln(x)=0\]
right?

Answer 5

yep, I'm following you so far

Answer 6

I think you can solve it by moving the ln(x) to the other side then exponentiating. I'm trying that now to make sure it works.

Answer 7

yeah I tried that before and got e^(x^4/6)

Answer 8

Hmm, I'm getting similar things...
I don't think there are other solutions for this one.
You may just have the one critical point.

Answer 9

Yeah, that's what I was thinking. Here, see if this link works, it's my webwork for this problem and it makes it look like there's 2 critical points. http://webwork2.morris.umn.edu/webwork2/F12CalcI_Mukherjee/4.3_Derivatives_and_graph_shape/5/?effectiveUser=bstottru&displayMode=jsMath&key=Lg6jDwbFQQXzWXFnt3KCwADcZv4z5dls&user=bstottru

Answer 10

I found A to be 0, which it said was right, but it wants me find B and that's where I'm confused.

Answer 11

Yeah, there is a vertical asymptote at x=0, and There should be a maximum before trending towards that horizontal asymptote . . .

Answer 12

Is my derivative wrong? Maybe that's the problem. I double checked it though and it seems fine.

Answer 13

I double-checked the derivative too and it is not fine..

Should be x^5 - 6x^5 ln(x) on top.

Answer 14

okay, just a sec

Answer 15

That should make things easier! :-)

Answer 16

ooohhhh yeah, whoops! okay, so let me solve for the numerator with the right equation this time....

Answer 17

Now it's lining up with the graph, and the critical point is a maximum.

Answer 18

so the other critical point should be "e"?

Answer 19

@CliffSedge

Answer 20

Not just e, e to a certain exponent.

Answer 21

oh wait, e^1/6

Answer 22

Okay, thanks for the help :)

Answer 23

There it is. Good job!