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!
Join our real-time social learning platform and learn together with your friends!