Question

What are the points of inflection of `f(x)=x/lnx`?

Answer 1

`x in{1,e^2}`

Explanation:

Find where `f''(x)=0` OR `f''(x)` switches from negative to positive instantaneously (like with an asymptote).

First, we find points where `f''(x)=0`.

`f'(x)=(lnx-x(1/x))/(lnx)^2=1/(lnx)-1/(lnx)^2`

`f''(x) = -1/(x(lnx)^2) +2/(x(lnx)^3)`

Now set this equal to 0.

`0 = -1/(x(lnx)^2) +2/(x(lnx)^3)`

`0 = -lnx+2`

`lnx = 2`

`x = e^2`

This shows that the only solution to `f''(x)=0` is `x=e^2`. Additionally, in order to be a point of inflection, the graph of `f''(x)` must cross the x-axis from positive to negative or negative to positive AT `x=e^2` (which can be proven by showing that `d/dx(x/lnx)!=0` when `x=e^2`, but I will not include this proof unless it is requested).

So, `x=e^2` is a point of inflection.

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We also need to look for points of instantaneous change. In this case, we have one: the asymptote at `x=1`.

At `x=1`, `ln(x)=0`, and it switches from negative to positive by definition. This means that `-1/(x(lnx)^2) +2/(x(lnx)^3)` will also jump from negative to positive, since the denominator is switching signs. Even though `f''(x)` never equals 0, it switches signs at `x=1`, so `x=1` is a point of inflection.

Therefore the two points of inflection for `f(x)=x/lnx` are `x=1` and `x=e^2`