This is an interesting problem. Thank you for posting it.

However, I feel there are a few things missing from the solution, which - if nothing more - increase a student's understanding and promote the application of necessary care when it comes to more complex problems.

1. It is important to clarify that the domain of f is the set of positive real numbers except 1.

2. Just after finding the local maximum at x = e, you write "Since f'(e) > 0", which is clearly an error; f'(e) = 0 and you have just located e as the solution to the equation f'(x) = 0.

3. It is not at all obvious (at least to me) that the graph that you show is the graph of f (apart from the discontinuity at x = 1). I think it would be better to invoke the second derivative test (a standard part of calculus). Calculate f''(x) = (1/x) . [(2-lnx)/(lnx)^3] and then f''(e) = 1/e > 0. From this, infer that f has a local minimum at x = e.

So, in the interval (1, +infinity), f(x) >= f(e) (*1)

Furthermore, for x > e:

lnx > lne or lnx > 1 of f'(x) > 0

Hence for x > e, f is strictly increasing and thus (*1) becomes a strict inequality. This means that f(x) > f(e) for x > e.

Since π > e, it follows that f(π) > f(e) or π/lnπ > e. QED

This is an interesting problem. Thank you for posting it.

ReplyDeleteHowever, I feel there are a few things missing from the solution, which - if nothing more - increase a student's understanding and promote the application of necessary care when it comes to more complex problems.

1. It is important to clarify that the domain of f is the set of positive real numbers except 1.

2. Just after finding the local maximum at x = e, you write "Since f'(e) > 0", which is clearly an error; f'(e) = 0 and you have just located e as the solution to the equation f'(x) = 0.

3. It is not at all obvious (at least to me) that the graph that you show is the graph of f (apart from the discontinuity at x = 1). I think it would be better to invoke the second derivative test (a standard part of calculus). Calculate f''(x) = (1/x) . [(2-lnx)/(lnx)^3] and then f''(e) = 1/e > 0. From this, infer that f has a local minimum at x = e.

So, in the interval (1, +infinity), f(x) >= f(e) (*1)

Furthermore, for x > e:

lnx > lne or

lnx > 1 of

f'(x) > 0

Hence for x > e, f is strictly increasing and thus (*1) becomes a strict inequality. This means that f(x) > f(e) for x > e.

Since π > e, it follows that f(π) > f(e) or π/lnπ > e. QED