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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?

See The Solution Submitted by Brian Smith    
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trying first part | Comment 1 of 2
F(x) should satisfy two conditions (to have a line tangent common at two different points x1 and x2):


The first condition implies F'(x1)-F'(x2)=0 which result in: 
4a*(x1^2+x1*x2+x2^2)+3b(x1+x2)+2c=0 [1]

To work out the second condition is useful:
  • x1^3-x2^3=(x1-x2)*(x1^2+x1*x2+x2^2). The bold expression can be worked from [1]
Anyway there is some calcutation involved, it's easy to get lost with potences and signs... At the end I obtain:
x1+x2=-c/2b [2]

and using [1]
x1*x2=(2ac^2-cb^2)/8ab^2 [3]

From [2] and [3] it's possibile to get x1 and x2 as an (a,b,c, dependent expression)

Apologize. Expresions [2] and [3] wrong...

Anyway: between x1 and x2 it would be an inflection point x3. For that point
F''(x3)=0   --> 12ax3^2+6bx3+2c=0 --> 
x3=(-3b +/- sq (9b^2-24ac)/12a which is real for 9b^2>24ac

With that condition we should have two inflection points (but only one of them betwen x1and x2) and there is a line tangent to F(x) in two different points.

Edited on April 15, 2016, 2:25 pm
  Posted by armando on 2016-04-15 06:38:13

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