Question

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

Answer 1

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`