Answer
(a) $d(x) =\sqrt {x^2+\frac{1}{x^2}}$
(b) See graph.
(c) $x=\pm1$
Work Step by Step
(a) $d(x)=\sqrt {(x)^2+(y)^2}=\sqrt {x^2+\frac{1}{x^2}}$
(b) See graph.
(c) Minima of $d(x)$ at $x=\pm1$
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