The set of positive real values of x for which the function f(x) = x/ln x is a decreasing function is ?

A x < e
B x = 1
C x < e^2
D x > e
E empty space

1 Answer
Dec 3, 2016

A #x< e#

Explanation:

#f(x) = x/lnx#

NB: #f(x)# is defined for #x in RR >0, x!=1#, so all values of x will be positive. Also, #f(x) -> -oo# as #x-> 1# from below and #f(x)-> +oo# as #x-> 1# from above.

To find a turning point set #f'(x) =0#

#f'(x) = (lnx * 1 - x*1/x)/[lnx]^2 = 0#

#lnx-1=0#

#lnx=1#

#x=e#

Hence #f(x)# has a turning point at #x=e#

Now observe the graph of #f(x)# below:

graph{x/lnx [-13.55, 17.64, -5.69, 9.9]}

It can be seen that #f(x)# is decreasing for #x < e# and increasing for #x>e#.

Hence the answer to this question is: A #x< e#