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

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

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

looks good to me

ok so now I have to find the critical numbers

thats right

how do I do that?

set the numerator = 0 and solve

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

why not the denominator? and I got lnx = 1

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

exactly

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

so do I ln the other side?

?

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

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

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

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

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

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

ohhh is x = e?

very good

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

oh man you got it :)

ok wait tho!

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

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

why is it (e, e) ?

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

got it thanks!

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