MHB Is f(x)=ln(x)/x Increasing or Decreasing?

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The function f(x) = ln(x)/x has a derivative of 1 - ln(x)/x^2, leading to critical points at (e, 1/e). The function is increasing on the interval (0, e) and decreasing on (e, positive infinity). There are no local minima, only a local maximum at (e, 1/e), and no inflection points or concavity. The analysis confirms the behavior of the function across its domain. Understanding these characteristics is essential for graphing and further analysis of f(x).
Emma_011
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Find where increasing/decreasing, concavity, local extrema and inflection points for f(x)=ln/x

So here is what I have so far:

The derivative is 1-ln(x)/x^2

Critical points are (e,1/e)

No concavity

Local max is also (e,1/e) (no local min)

no inflection points

Increase on (0, e) and decrease on (e, positive infinity)

Is this correct? I tried to do a graph to justify my work
 
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Emma_011 said:
Find where increasing/decreasing, concavity, local extrema and inflection points for f(x)=ln x/x

The derivative is 1-ln(x)/x^2
Critical points are (e,1/e)
No concavity
Local max is also (e,1/e) (no local min)
Hi Emma, ☺

All correct so far.
Btw, how did you conclude that there is no concavity?

no inflection points
How did you reach that conclusion?
What is an inflection point exactly?

Increase on (0, e) and decrease on (e, positive infinity)

Correct.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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