Question 909355
your equation is y = 1/x


this is a graph of your equation.


<img src = "http://theo.x10hosting.com/2014/100501.jpg" alt="$$$" </>


p is a point on the graph of y = 1/x.


d = the distance from p to the origin.


on the graph, p is represented by y which is the result of the equation of y = 1/x for any value of x.


since the equation is y = 1/x, then you want to find the formula for the distance from any point on that graph  to the origin.


the distance from any point on the graph to the origin is given by the generic equation of {{{d = sqrt(x^2 + y^2)}}}


when y = 1/x, this equation becomes {{{d = sqrt(x^2 + (1/x)^2)}}} which becomes:


{{{d = sqrt((x^4+1)/x^2)}}} which becomes:


{{{d = sqrt(x^4+1)/x}}}.


you want to graph {{{d = sqrt(x^4+1)/x)}}}


to graph it, replace d with y to get {{{y = sqrt(x^4+1)/x)}}}


the graph of that equation is shown below.


<img src = "http://theo.x10hosting.com/2014/100502.jpg" alt="$$$" </>


the particular software i am using marks the max/min points on each curve of the graph.


it is showing me that the minimum point if at x = 1 and the maximum point is at x = -1.


it is also showing me that the distance to the origin will be 1.41414 at that point, which is the samee as sqrt(2).


the minimum distances from any point on the graph of y = 1/x to the origin will occur when x = -1 and when x = 1.


when x = -1, y = -1, and when x = 1, y = 1.


the distance to the origin in both cases will be {{{d = sqrt(x^2 + y^2)}}} which becomes {{{d = sqrt(1^2 + 1^2)}}} which becomes {{{d = sqrt(2)}}}, or {{{d = sqrt((-1)^2 + (-1)^2)}}} which still becomes {{{d = sqrt(2)}}}.


the solution to your problem is that d is the smallest when x = -1 and when x = 1.


the particular calculator that i used is at:


<a href = "http://www.desmos.com/calculator" target="_blank">http://www.desmos.com/calculator</a>