All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math > Calculus
Quartic Questions (Posted on 2016-04-13) Difficulty: 3 of 5
Let F(x) be the fourth degree polynomial ax^4 + bx^3 + cx^2 + dx + e, with a>0.

When does F(x) have a line which is tangent to it at two different points?

When does F(x) have two distinct inflection points?

  Submitted by Brian Smith    
Rating: 4.0000 (1 votes)
Solution: (Hide)
Question 1:
Express F(x) as the sum of a squared quadratic minus a line. The line will be the tangent line and the roots of the quadratic are the x coordinate locations of the two tangent points.
F(x) = ([sqrt(a)]*x^2 + [b/(2*sqrt(a))]*x + [(4ac-b^2)/(8*sqrt(a^3))])^2 - ([-d+b*(4ac-b^2)/(8a^2))]*x + [-e+b^2*(4ac-b^2)^2/(64a^3)])

For F(x) to have a double tangent the quadratic must have two real roots, that means its discriminant must be greater than 0:
[b/(2*sqrt(a))]^2 - 4*[sqrt(a)]*[(4ac-b^2)/(8*sqrt(a^3))] > 0
This simplifies to (3b^2 - 8ac)/(4a) > 0. Since A is nonzero then 3b^2 - 8ac > 0.

Question 2:
The second derivative of F(x) is F"(x) = 12ax^2 + 6bx + 2c. To have two distinct inflection points the second derivative must have two distinct real roots. Since F"(x) is quadratic this implies its discriminant must be greater than 0:
(6b)^2 - 4*(12a)*(2c) > 0.
This simplifies to 3b^2 - 8ac > 0.

In both cases the necessary condition is 3b^2 - 8ac > 0. This implies the two conditions are equivalent.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Colombus eggarmando2016-04-16 17:40:05
trying first partarmando2016-04-15 06:38:13
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (23)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information