Critical Points of a Function: Finding the Points of Inflection

Click For Summary

Homework Help Overview

The discussion revolves around finding critical points of the function g(x) = x - (5/x^2) and g(x) = x |x + 5|. Participants are exploring the conditions under which critical points occur, specifically focusing on where the derivative is zero or undefined.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the process of taking the derivative and setting it to zero to find critical points. There are questions about the conditions under which the derivative is undefined and how that relates to the critical points. Some participants also question the interpretation of critical points in relation to the function's domain.

Discussion Status

There is an ongoing exploration of the definitions and conditions for critical points, with some participants providing guidance on identifying where the derivative is zero and where it is undefined. Multiple interpretations of critical points are being discussed, particularly regarding the relationship between the function's domain and the existence of critical points.

Contextual Notes

Participants are considering the implications of the function being undefined at certain points and how that affects the identification of critical points. There is a reference to specific functions and their behavior at critical points, highlighting the nuances in defining critical points based on the function's domain.

Niaboc67
Messages
249
Reaction score
3

Homework Statement



1.) Let g(x) = x − (5/x^2)

find all critical numbers (if any) of g.

Give answers in increasing order (smallest first). Enter DNE in any unused space.

x =

x =

and

2.) Let g(x) = x |x + 5|

The Attempt at a Solution

[/b]

I know I must first take the derivative and then set to zero, in order to find it's zero.

g(x) = x - 5/x^2
g'(x) = 1+10x^-3
g'(x) = 10/x^3 = -1
g'(x) = (x^3)10/x^-3 = -1(x^3)
g'(x) = 10 = -1(x^3)
g'(x) = -10 = x^3
g'(x) = -10^(1/3) = x
After here I don't know what I am suppose to do in order to find the crical points :/
 
Last edited by a moderator:
Physics news on Phys.org
Critical points are where the derivative is zero or undefined. You've found where it's zero. Now, for what value(s) of x is it undefined?
 
  • Like
Likes   Reactions: RJLiberator
That seems to be right. You have a correct x value from my understanding. To find the point, you input x into the correct function to receive the y output.

edit: Andrew posted a better response.
 
Niaboc67 said:

Homework Statement


1.) Let g(x) = x − (5/x^2)

find all critical numbers (if any) of g.

Give answers in increasing order (smallest first). Enter DNE in any unused space.
x =
x =
and

2.) Let g(x) = x |x + 5|

The Attempt at a Solution


[ B ]
I know I must first take the derivative and then set to zero, in order to find it's zero.

g(x) = x - 5/x^2
g'(x) = 1+10x^-3
That is correct for the derivative, g'(x) .

After this you apparently meant to set g'(x) = 0 and then solve for x. However, you kept setting each line equal to g'(x).

None of the following lines is equal to g'(x).
Niaboc67 said:
g'(x) = 10/x^3 = -1
g'(x) = (x^3)10/x^-3 = -1(x^3)
g'(x) = 10 = -1(x^3)
g'(x) = -10 = x^3
g'(x) = -10^(1/3) = x
After here I don't know what I am suppose to do in order to find the critical points :/[ /B ]
There are many contexts for critical numbers.

Here it appears that you are trying to find places at which the derivative can change sign.

Essentially, you need to identify places where the derivative is zero and places where the derivative is undefined.
 
  • Like
Likes   Reactions: vela
andrewkirk said:
Critical points are where the derivative is zero or undefined. You've found where it's zero. Now, for what value(s) of x is it undefined?

Does that definition (non-existence of derivative) apply also at points where the function itself is undefined? That is the case here!
 
SammyS said:
None of the following lines is equal to g'(x).
Niaboc67 said:
g'(x) = 10/x^3 = -1
g'(x) = (x^3)10/x^-3 = -1(x^3)
g'(x) = 10 = -1(x^3)
g'(x) = -10 = x^3
g'(x) = -10^(1/3) = x
After here I don't know what I am suppose to do in order to find the critical points :/
To elaborate on what SammyS said, the last correct line you (Niaboc67) wrote before the first line quoted above was
g'(x) = 1+10x^-3

The next step should be to set g'(x) = 0
##g'(x) = 0 \Rightarrow 1 + 10x^{-3} = 0##
Now, work with that.
 
Ray Vickson said:
Does that definition (non-existence of derivative) apply also at points where the function itself is undefined? That is the case here!
I presume so, as the function being undefined at ##x=a## necessarily implies that the derivative is undefined there, even if ##\lim_{x\to a}f'(x)## exists.

To me it seems more natural to define a critical point as one where the derivative is zero or the function is undefined. A difference would be cases like ##y=x^\frac{1}{3}##, which would be critical at at ##x=0## under the first definition but not under that alternative.
 
The critical point must belong to the domain of the function. Think of the function g(x)=√x. Are all x<0 critical points? x=0 is the only critical point, as the function exists there but its derivative does not exist.

We say that [PLAIN]http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints_files/eq0001M.gif[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints_files/empty.gif is a critical point of the function f(x) if f(x) exists and if either of the following are true: f '(c) =0 or f '(c) does not exist. [PLAIN]http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints_files/empty.gif[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints_files/empty.gif
 
Last edited by a moderator:

Similar threads

  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
2K
  • · Replies 28 ·
Replies
28
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K