Question

calculator question about relative maximum, relative minimum and point of inflection

Answer 1

\[f \prime=\cos(lnx)\]

Answer 2

I am given that first derivative of a function
f'(x)=cos(lnx)
I am suppose to identify relative maximum
x=4.8104774
I am suppose to find the relative minimum
x=.20787958
I am not sure about the point of inflection
I took the second derivative and solved for 0
I think I is x=1

Answer 3

Did you get this for your second derivative?\[\Large\rm f''(x)=-\frac{1}{x}\sin(\ln x)\]

Answer 4

I just cheated and put it in my calculator

Answer 5

lol :)

Answer 6

the TI calculator just takes the general command and will create the 2nd and even the third derivative graph of any function.

Answer 7

That doesn't seem like a good approach -_- hmmm

Answer 8

It really does work

Answer 9

I just put in the 2nd derivative in the graphing calculator
ie
y1=1st derivative function given
y2=2nd derivative command (it is Math 8) on the graphing calculator
I am not sure if I was to locate the zero value and that would be the point of inflection
if I have a sign change

Answer 10

I'm just.. I don't know how to help you if you're just using your calculator :\
It's just weird..

Answer 11

That's ok

Answer 12

x=1 seems correct.
Looks like there is another weird point though.. hmmm

Answer 13

https://www.desmos.com/calculator/s79b4blgj6

Answer 14

See that other critical point?
That sharp dip downward near zero?

It's weird too, cause it didn't show up when I did the calculations by hand.. hmm

Answer 15

yes and that is the graph I got with my calculator and that is how I located the min and max

Answer 16

I am able to request the next derivative using my graphing calculator

Answer 17

It'll graph the second derivative for you? oo that should help.

Answer 18

Oh darn, it looks kind of crazy though doesn't it? :P