• Support PF! Buy your school textbooks, materials and every day products Here!

Finding points of inflection

1. Homework Statement
Suppose the graph f''(x) of a function is given by:
(see attachment)

(a) Find all points of inflection of f(x)

3. The Attempt at a Solution
First I figured, by looking at the graph and seeing the intercept points [(-2,0),(0,0),(2,0)] that f''(x) = [tex]x^{4}-4x^{2}[/tex]
solving for f''(x)=0 gives us x = -2 or 0 or 2.
Between -2 < x < 2 there is no change of sign, which indicates no point of inflection.
Thus the points of inflection for f(x) are at x = -2 and x = 2.
correct?
But how can I find the y intercepts with just what I have? I only know how to find them by putting x back into f(x). As we don't have f(x) here, I'm stuck. Integrating back doesn't help because of the unknown variables.
Or am I trying too hard, and what I've done is the answer?
 

Attachments

Answers and Replies

Gib Z
Homework Helper
3,344
4
Well, remember that an inflection point of f(x) is where the first derivative is at either a maximum or minimum. And find first the maxima and minima of the first derivative, we must solve the zeros of the third derivative =D
 
HallsofIvy
Science Advisor
Homework Helper
41,738
897
1. Homework Statement
Suppose the graph f''(x) of a function is given by:
(see attachment)

(a) Find all points of inflection of f(x)

3. The Attempt at a Solution
First I figured, by looking at the graph and seeing the intercept points [(-2,0),(0,0),(2,0)] that f''(x) = [tex]x^{4}-4x^{2}[/tex]
Okay, that looks good.

solving for f''(x)=0 gives us x = -2 or 0 or 2.
?? If f(x)= x4 - 4x2 then f '(x)= 4x3- 8x and f "(x)= 12x2- 8. That is not 0 at x= 0, 2, and -2!

Between -2 < x < 2 there is no change of sign, which indicates no point of inflection.
Thus the points of inflection for f(x) are at x = -2 and x = 2.
correct?
But how can I find the y intercepts with just what I have? I only know how to find them by putting x back into f(x). As we don't have f(x) here, I'm stuck. Integrating back doesn't help because of the unknown variables.
Or am I trying too hard, and what I've done is the answer?
 
third derivative is [tex]4x^{3}-8x[/tex]
solving for f'''(x) = 0 gives us
x = +/-[tex]\sqrt{2}[/tex] and x = 0.

minima points of f''(x) are at x = (-[tex]\sqrt{2}[/tex], -4) & ([tex]\sqrt{2}[/tex], -4)
maxima point of f''(x) is (0, 0).

I still confused here! How does this help me find the y co-ordinates for inflection points of f(x)?
 
?? If f(x)= x4 - 4x2 then f '(x)= 4x3- 8x and f "(x)= 12x2- 8. That is not 0 at x= 0, 2, and -2!
the graph x4 - 4x2 is the second derivative, not f(x).
 

Related Threads for: Finding points of inflection

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
959
Replies
19
Views
779
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
5
Views
1K
Replies
3
Views
5K
Top