F(x) should satisfy two conditions (to have a line tangent common at two different points x1 and x2):
F'(x1)=F'(x2)
and
F(x1)-F(x2)/x1-x2=F'(x1)=F'(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
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Posted by armando
on 2016-04-15 06:38:13 |
trying first part
