Question

find the absolute extrema of g(x) = x/lnx on [2,5]

Answer 1

what you wanna do is find the derivative of x/ln x, first

Answer 2

for the derivative I got (lnx - 1) / (lnx)^2 is that right?

Answer 3

looks good to me

Answer 4

ok so now I have to find the critical numbers

Answer 5

thats right

Answer 6

how do I do that?

Answer 7

set the numerator = 0 and solve

Answer 8

that will give you the critcal points, but also keep in mind that since this is on a closed interval, 2 and 5 must also be evluated

Answer 9

why not the denominator? and I got lnx = 1

Answer 10

what lagrange said. you have to check
\[\frac{2}{\ln(2)},\frac{5}{\ln(5)}\] and your critical point

Answer 11

exactly

Answer 12

you got
\[\ln(x)=1\] you need x

Answer 13

so do I ln the other side?

Answer 14

?

Answer 15

you dont have to ln anything. All you have to do is find the value of the critcal point and 2 and 5. The you loswest value is the abs min and you highest value is the abs max. Anthing else is rel. extrema

Answer 16

but what do I do with ln(x) = 1? I need to find the critical point

Answer 17

and i don't understand how to test 2 and 5

Answer 18

you can evluate 2 and 5 by simply plugging them into the orginal function

Answer 19

ok so i checked 2 and 5. but i still don't have a critical point to see if it is a max or min

Answer 20

ok is it clear what x is if
\[ln(x)=1\]?

Answer 21

ohhh is x = e?

Answer 22

very good

Answer 23

so if e = 2.718, then that is a minimum. and 5/ln5 is a maximum

Answer 24

oh man you got it :)

Answer 25

ok wait tho!

Answer 26

the absolute max is (5, ln5) and the min is at (e, ?)

Answer 27

the abs. max at (5, (5/ln5)) and abs min at (e, e)

Answer 28

why is it (e, e) ?

Answer 29

because that when the derivative is zero, and so when you evluate it you get e

Answer 30

got it thanks!