/sci/ - Science & Math » Thread #10881924

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Downtime Math #9

Anonymous Mon 12 Aug 08:43:48 2019 No.10881924 View Reply Original Report

Quoted By: >>10882937 >>10882943 >>10884673 >>10885930

Hi /sci/
Last question was solved pretty well.
Hopefully this one interests you too.
Last post: >>10879442

Anonymous

Anonymous Mon 12 Aug 2019 20:06:41 No.10882937 Report

Quoted By: >>10882956

>>10881924
g(x) = (a x^2 + b x + c) e^(-x)
g'(x) = (-a x^2 + (2a - b) x + (b - c)) e^(-x)

0 = g'(1) = (a - c) e^(-1)
c = a
0 = g'(4) = (-9a - 3b) e^(-4)
b = -3a

g(x) = a (x^2 - 3 x + 1) e^(-x)

For the tangent condition the equation
g'(x) = (g(x) - k)/x
must have exactly 3 solutions for -1 < k < 0
which only happens for
$ a = \frac{\sqrt{13} - 3}{26} e^{\frac{\sqrt{13} + 7}{2}} $

So
$ g(x) = \frac{\sqrt{13} - 3}{26} e^{\frac{\sqrt{13} + 7}{2}} \left( x^2 - 3 x + 1 \right) e^{-x} $
$g(-2) \times g(4) = \frac{55}{338} \left( 11 - 3 \sqrt{13} \right) e^{5 + \sqrt{13}} \approx 163 $

Anonymous

Anonymous Mon 12 Aug 2019 20:10:11 No.10882943 Report

Quoted By: >>10884676

>>10881924
0 by symmetry

Anonymous

Anonymous Mon 12 Aug 2019 20:17:02 No.10882956 Report

Quoted By:

>>10882937
The sign of $a$ is wrong.

Anonymous

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Anonymous Tue 13 Aug 2019 13:48:48 No.10884659 Report

Quoted By:

The answer is 72

Anonymous

Anonymous Tue 13 Aug 2019 13:56:37 No.10884673 Report

Quoted By:

>>10881924
don't know what x in g(-2) x g(4) means

Anonymous

Anonymous Tue 13 Aug 2019 13:58:41 No.10884676 Report

Quoted By:

>>10882943
This meme is so weird.

Anonymous

Anonymous Wed 14 Aug 2019 00:36:43 No.10885930 Report

Quoted By:

>>10881924
Hey, OP,

I managed to get the solution of 72*b^2*e^(-2) where b is a constant greater than (-27*k*e^x)/32 for some fixed number -1<k<0.

Am I close?

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