Extrema and Inflection Point

In summary: 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 dealing with functions of several variables).
  • #1
204
7

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.
 
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  • #2
RyanTAsher said:

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##.
 
  • #3
Ray Vickson said:
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?

Concavity.jpg
 
  • #4
  • #5
Yes, that is exactly what I mean. Thank you for your time!
 
  • #7
SammyS said:
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.
 
  • #8
Ray Vickson said:
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.
 
  • #9
SammyS said:
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.
 
  • #10
Ray Vickson said:
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).
 
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  • #11
Mark44 said:
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.
 
  • #12
Ray Vickson said:
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.
 

1. What is an extremum?

An extremum is a point on a graph where the value of a function is either at its highest (maximum) or lowest (minimum) point. It can also refer to the highest or lowest value of a function over a certain interval.

2. How do you find extrema on a graph?

To find extrema on a graph, you can use the first derivative test. First, find the critical points by setting the first derivative equal to zero and solving for x. Then, use the second derivative test to determine whether each critical point is a maximum or a minimum.

3. What is an inflection point?

An inflection point is a point on a graph where the curvature changes from concave up to concave down, or vice versa. It is where the second derivative of a function changes sign.

4. How are extrema and inflection points related?

Extrema and inflection points are related because they both involve the behavior of a function with respect to its derivatives. Extrema occur at points where the first derivative is zero, while inflection points occur at points where the second derivative is zero.

5. Why are extrema and inflection points important?

Extrema and inflection points are important because they provide information about the behavior of a function. They can help determine the maximum and minimum values of a function, as well as the direction of its curvature. This information can be useful in many applications, such as optimization problems and curve sketching.

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