Question 1127347
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The measure of the segment *[tex \Large AC] where *[tex \Large A] is the point *[tex \Large (-1,6)] and *[tex \Large C] is an arbitrary point *[tex \Large x]-axis, *[tex \Large (x,\,0)] is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ AC\ =\ \sqrt{(-1\,-\,x)^2\ +\ (6\ -\ 0)^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ AC\ =\ \sqrt{x^2\ +\ 2x\ +\ 37}]


Similarly,  the measure of *[tex \Large CB] is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ CB\ =\ \sqrt{(x\,-\,14)^2\ +\ (9\ -\ 0)^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ CB\ =\ \sqrt{x^2\ -\ 28x\ +\ 277}]


The total distance is the sum of these two expressions, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d(x)\ =\ \sqrt{x^2\ +\ 2x\ +\ 37}\ +\ \sqrt{x^2\ -\ 28x\ +\ 277}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d'(x)\ =\ \frac{2x\ +\ 2}{2\sqrt{x^2\ +\ 2x\ +\ 37}}\ +\ \frac{2x\ -\ 28}{2\sqrt{x^2\ -\ 28x\ +\ 277}}]


Set the first derivative equal to zero and solve for the critical values.  Make sure to discard any extraneous roots.  Then evaluate *[tex \LARGE d(x)] for that critical value.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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