# Extrema and Inflection Point

## Homework Statement

Find all extrema and inflection points of the function ## y = \frac {x} {ln(x)} ##

## Homework Equations

I did the first and second derivative by hand, and they worked out in CAS as well...

## y = \frac {x} {ln(x)} ##
## y' = \frac {lnx - 1} {(ln(x))^2} ##
## y'' = \frac {2 - ln(x)} {x(ln(x))^3} ##

## The Attempt at a Solution

EXTREMA:

## y' = \frac {lnx -1} {(ln(x))^2} = 0 ##
## ln(x) - 1 = 0 ##
## ln(x) = 1 ##
## x = e ##

.....-.............+......
<-------- e -------->

## y = \frac {e} {ln(e)} ##
## y = e ##

Therefore, a minima exists at ##(e,e)##

INFLECTION POINT:

## y'' = \frac {2 - ln(x)} {x(ln(x))^3} = 0 ##
## y'' = 2 - ln(x) = 0 ##
## ln(x) = 2 ##
## x = e^2 ##

....+.............-......
<------ e^2 ------>

## y = \frac {e^2} {ln(e^2)} ##
## y = \frac {e^2} {2} ##

Therefore, an inflection point exists at ## (e^2,\frac {e^2} {2}) ##

===============================================

Looking at the graph of ## y = \frac {x} {lnx} ##, the extrema is correct.

Also, I noticed by solving for the denominators = 0, I find that there is a vertical asymptote at x = 1, which seems like it should be the inflection point... not ## (e^2,\frac {e^2} {2}) ##, as the graphing is just continually concave up at ##e^2##, not changing concavity.

I also know the domain is ##(0,1)##U##(1,\infty)## due to the denominator, and property of ##ln(x)##.

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EDIT:

I think I may have gotten it... I forgot to use the critical point for ## y'' = \frac {2 - ln(x)} {x(ln(x))^3}##, when the bottom is equal to zero, which you solve analytically.

##x(ln(x))^3) = 0,## when ## x = 1##

So this tells us there is an inflection point at 1, but what does this mean of the ##e^2## result, I don't see how that is invalid, yet it doesn't satisfy the equation.

Last edited:

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Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Find all extrema and inflection points of the function ## y = \frac {x} {ln(x)} ##

## Homework Equations

I did the first and second derivative by hand, and they worked out in CAS as well...

## y = \frac {x} {ln(x)} ##
## y' = \frac {lnx - 1} {(ln(x))^2} ##
## y'' = \frac {2 - ln(x)} {x(ln(x))^3} ##

## The Attempt at a Solution

EXTREMA:

## y' = \frac {lnx -1} {(ln(x))^2} = 0 ##
## ln(x) - 1 = 0 ##
## ln(x) = 1 ##
## x = e ##

.....-.............+......
<-------- e -------->

## y = \frac {e} {ln(e)} ##
## y = e ##

Therefore, a minima exists at ##(e,e)##

INFLECTION POINT:

## y'' = \frac {2 - ln(x)} {x(ln(x))^3} = 0 ##
## y'' = 2 - ln(x) = 0 ##
## ln(x) = 2 ##
## x = e^2 ##

....+.............-......
<------ e^2 ------>

## y = \frac {e^2} {ln(e^2)} ##
## y = \frac {e^2} {2} ##

Therefore, an inflection point exists at ## (e^2,\frac {e^2} {2}) ##

===============================================

Looking at the graph of ## y = \frac {x} {lnx} ##, the extrema is correct.

Also, I noticed by solving for the denominators = 0, I find that there is a vertical asymptote at x = 1, which seems like it should be the inflection point... not ## (e^2,\frac {e^2} {2}) ##, as the graphing is just continually concave up at ##e^2##, not changing concavity.

I also know the domain is ##(0,1)##U##(1,\infty)## due to the denominator, and property of ##ln(x)##.

======
======

EDIT:

I think I may have gotten it... I forgot to use the critical point for ## y'' = \frac {2 - ln(x)} {x(ln(x))^3}##, when the bottom is equal to zero, which you solve analytically.

##x(ln(x))^3) = 0,## when ## x = 1##

So this tells us there is an inflection point at 1, but what does this mean of the ##e^2## result, I don't see how that is invalid, yet it doesn't satisfy the equation.
On ##(1,\infty)## ##f(x) = x/\ln(x)## is convex for ##x < e^2## and concave for ##x > e^2##. The function has the rather un-intuitive property that ##f(x) \to +\infty## but ##f'(x) \to 0## as ##x \to \infty##.

On ##(1,\infty)## ##f(x) = x/\ln(x)## is convex for ##x < e^2## and concave for ##x > e^2##. The function has the rather un-intuitive property that ##f(x) \to +\infty## but ##f'(x) \to 0## as ##x \to \infty##.
Okay, so, on this graph I've made, is that point at (e^2, e^2/2) actually an inflection point, but it is just so subtle that it looks like it is still concave up to the human eye at this zoom?

Ray Vickson
Homework Helper
Dearly Missed
Yes, that is exactly what I mean. Thank you for your time!

Ray Vickson
Homework Helper
Dearly Missed
The terminology "concave up" and "concave down" is very commonly used in introductory Calculus textbooks.
I am aware of that, but there is nothing wrong with exposing the OP to the more modern terminology. Of course, when writing up his solution for marks he would be wise to use the same terminology as his instructor or textbook.

SammyS
Staff Emeritus
Homework Helper
Gold Member
I am aware of that, but there is nothing wrong with exposing the OP to the more modern terminology. Of course, when writing up his solution for marks he would be wise to use the same terminology as his instructor or textbook.
I did not realize that it was more modern terminology.

Ray Vickson
Homework Helper
Dearly Missed
I did not realize that it was more modern terminology.
Both types are widely used still. However, I believe that the terms "concave up/down" appear mostly in calculus textbooks, having been used first in the 19th century and then copied over and over again ever since. Many more advanced treatments (especially in applications such as optimization, etc.) use the newer terminology, which becomes particularly important when one discusses generalizations such as pseudo-convex, quasi-convex, K-convex and the like (with similar names when 'convex' is replaced by 'concave'). These would be horribly ungainly if they were referred to as pseudo-concave-up, etc.

Mark44
Mentor
Both types are widely used still. However, I believe that the terms "concave up/down" appear mostly in calculus textbooks, having been used first in the 19th century and then copied over and over again ever since.
Notwithstanding the terms pseudo-convex, quasi-convex, and so on, I believe the terms "concave up" and concave down" are clearer and thereby more easily understood.

If you look at the graph of y = x2, it is obviously concave up. If you omit the direction (up or down), the graph is concave from the perspective of a viewpoint below the graph, but is convex, from the perspective of a viewpoint above the graph. Put another way, the graph of this function divides the plane into two regions: one that is concave (the portion below the graph), and one that is convex (the region above the graph).

Ray Vickson
Ray Vickson
Homework Helper
Dearly Missed
Notwithstanding the terms pseudo-convex, quasi-convex, and so on, I believe the terms "concave up" and concave down" are clearer and thereby more easily understood.

If you look at the graph of y = x2, it is obviously concave up. If you omit the direction (up or down), the graph is concave from the perspective of a viewpoint below the graph, but is convex, from the perspective of a viewpoint above the graph. Put another way, the graph of this function divides the plane into two regions: one that is concave (the portion below the graph), and one that is convex (the region above the graph).
I have never actually heard of a "concave region.

However, even though the function is "concave up" the "upper" region (above the graph) is convex. Oh well.

Mark44
Mentor
I have never actually heard of a "concave region.
Maybe not that term, exactly, but it's descriptive enough that you should get the idea I'm trying to convey.
Ray Vickson said:
However, even though the function is "concave up" the "upper" region (above the graph) is convex. Oh well.