Find the absolute extrema of the function on the interval [2,3]. (Round your answers to the nearest hundredth.)
Absolute Minimum ___ at x=___
Absolute Maximum ___ at x=___
4 answers
I am sorry, the function is g(x)= x/(ln(x))
g' = (lnx-1)/ln^2(x)
since any local max/min occur at ln(x) = 1, that is at x=e
So, if that's a local max, then it is the interval max on [2,3].
g'' = (2-lnx)/(x ln^3(x))
g''(e) = 1/e > 0, so the graph is concave up at x=e, making f(e) a minumum.
So, max on [2,3] is the greater of f(2) and f(3).
f(2) = 2.88
f(3) = 2.73
so, the max on the interval [2,3] = 2/ln2
The minimum is f(e) = e = 2.72
since any local max/min occur at ln(x) = 1, that is at x=e
So, if that's a local max, then it is the interval max on [2,3].
g'' = (2-lnx)/(x ln^3(x))
g''(e) = 1/e > 0, so the graph is concave up at x=e, making f(e) a minumum.
So, max on [2,3] is the greater of f(2) and f(3).
f(2) = 2.88
f(3) = 2.73
so, the max on the interval [2,3] = 2/ln2
The minimum is f(e) = e = 2.72
dg/dx = [ ln x (1) - x (1/x) ] / (ln x)^2
= [ln x - 1 ] / (ln x)^2
that is zero when x = e or x = about 2.72
is that a max or a min?
you could take another derivative, but easier to check when x = 2.5 or something
if x = 2.5
dg/dx = [.916 -1 ] / .84 = negative
so as x goes from 2.5 to 2.72 the derivative goes from negative to zero
therefore there is a minimum at x = e
calculate it
now check the value of g at the end points, x = 2 and x = 3 and see which is bigger
= [ln x - 1 ] / (ln x)^2
that is zero when x = e or x = about 2.72
is that a max or a min?
you could take another derivative, but easier to check when x = 2.5 or something
if x = 2.5
dg/dx = [.916 -1 ] / .84 = negative
so as x goes from 2.5 to 2.72 the derivative goes from negative to zero
therefore there is a minimum at x = e
calculate it
now check the value of g at the end points, x = 2 and x = 3 and see which is bigger
Thanks to both of you, I won't forget how to do this next time! It makes much more sense now, and will be even easier on basic functions versus logarithmic/exponential functions. I think it was the natural log that tripped me a little bit there, but thank you for the explanations!