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OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

looks good to me

OpenStudy (anonymous):

ok so now I have to find the critical numbers

OpenStudy (anonymous):

thats right

OpenStudy (anonymous):

how do I do that?

OpenStudy (anonymous):

set the numerator = 0 and solve

OpenStudy (anonymous):

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

OpenStudy (anonymous):

why not the denominator? and I got lnx = 1

OpenStudy (anonymous):

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

OpenStudy (anonymous):

exactly

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so do I ln the other side?

OpenStudy (anonymous):

?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ohhh is x = e?

OpenStudy (anonymous):

very good

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oh man you got it :)

OpenStudy (anonymous):

ok wait tho!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

why is it (e, e) ?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

got it thanks!

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