Max points, points of inflection

  • Thread starter Thread starter fitz_calc
  • Start date Start date
  • Tags Tags
    Max Points
Click For Summary
The discussion focuses on finding the maximum and minimum points, as well as the inflection points of the function y=4x^3 - 3x^4. A participant initially identifies two local maxima but is corrected, as the textbook indicates there is only one maximum point. The importance of verifying critical points as local extrema is emphasized, along with the need to clearly define inflection points. The second derivative is questioned, suggesting discrepancies in the calculations of inflection points. Clarifying these concepts and calculations is essential for accurate results.
fitz_calc
Messages
41
Reaction score
0

Homework Statement



which best describes y=4x^3 - 3x^4
find max/min points and inflection points

The Attempt at a Solution



When I work this one out I get x<0 and 0<x<1 as my two local max points. However, the book says there is only ONE max point - why is this?

with the second derivative I do get two inflection points of 0 and 1/2 which I assume to be correct.

Thoughts??
 
Physics news on Phys.org
I think that you'll see where your mistakes are once you write down a clear definition of an inflection point, and of a local extremum. I'll go ahead and say this (because it's often a step that students forget): did you check that your critical points are in fact local max/min points? Likewise for inflection points.

Can you also post your solution?
 
fitz_calc said:

Homework Statement



which best describes y=4x^3 - 3x^4
find max/min points and inflection points

The Attempt at a Solution



When I work this one out I get x<0 and 0<x<1 as my two local max points. However, the book says there is only ONE max point - why is this?
That makes no sense at all. Max/min points are individual points, not sets of points. HOW did you "work this one out"? If you mean that you got x= 0 and x= 1 as your max/min points, it is true that the derivative is 0 at x= 0 and x= 1, but that is not enough to be a max or a min.

with the second derivative I do get two inflection points of 0 and 1/2 which I assume to be correct.
That is not at all what I get. What is the second derivative?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
2K
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
998
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
12
Views
1K