Question 1208130
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Part A


(x1,y1) = origin = (0,0)
(x2,y2) = point P = (x,y) = (x,1/x) since y = 1/x


Distance formula from (x1,y1) to (x2,y2)
{{{d = sqrt( (x2-x1)^2 + (y2-y1)^2 ) }}}


{{{d = sqrt( (x-0)^2 + (y-0)^2 ) }}}


{{{d = sqrt( x^2 + y^2 ) }}}


{{{d(x) = sqrt( x^2 + (1/x)^2 ) }}}
Optionally you can simplify this, but I'll let the student take over from here.
Note that the distance formula is based on the Pythagorean Theorem.


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Part B


This is what the graph of d(x) looks like
{{{graph(400,400,-5,5,-5,5,-100,sqrt( x^2 + (1/x)^2 ))}}}
There are many graphing tools to pick from. 
My favorite is GeoGebra, but Desmos is really good as well. 
If you are familiar with something like a TI83, then it might be best to stick to that.


Things to pay attention to:<ul><li>x = 0 isn't allowed since it causes a division by zero error. This explains the vertical asymptote at x = 0.</li><li>The curve is symmetric about the y axis. The function d(x) is even. We can prove this algebraically by showing that d(-x) = d(x) for all x in the domain. I'll leave the proof for the reader.</li><li>The smallest distance possible is d = sqrt(2) = 1.41421356 approximately, which occurs when either x = 1 or x = -1. Use calculus to find this, or use the graphing calculator's minimum locator feature.</li></ul>
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