Critical Points of a Function: Finding the Points of Inflection

In summary: 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.
  • #1
Niaboc67
249
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 :/
 
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  • #2
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?
 
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  • #3
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.
 
  • #4
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.
 
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  • #5
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!
 
  • #6
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.
 
  • #7
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.
 
  • #8
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
 
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Related to Critical Points of a Function: Finding the Points of Inflection

1. What are critical points of a function?

Critical points of a function are the points where the derivative of the function is equal to zero or does not exist. These points are important because they can indicate the locations of maximum or minimum values of the function.

2. How do you find critical points of a function?

To find critical points of a function, you need to take the derivative of the function and set it equal to zero. Then, solve for the variable to find the x-values of the critical points. You may also need to check for any points where the derivative does not exist.

3. What is the significance of critical points in calculus?

Critical points are significant in calculus because they can help us determine the maximum and minimum values of a function. This information is useful in optimization problems, where we want to find the most efficient or optimal solution to a given problem.

4. Can a function have more than one critical point?

Yes, a function can have multiple critical points. For example, a cubic function can have up to three critical points. These points may represent local maximum or minimum values, or they may be inflection points where the concavity of the function changes.

5. How do critical points relate to the graph of a function?

The critical points of a function are the points where the slope of the graph is either zero or undefined. This means that the graph will have a horizontal tangent line or a vertical tangent line at these points. They can also mark the locations of maximum or minimum values on the graph.

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