Answer
(a) $d(x)=\sqrt{x^2+\frac{1}{x^2}}$
(b) Refer to th graph below.
(c) $d$ is smallest when $x=-1, 1$.
Work Step by Step
(a) The distance $d$ from $P(x,y)$ to the origin$(0,0)$ is:
$$\begin{align*}
d&=\sqrt{(x-x_1)^2+(y-y_1)^2}\\
&=\sqrt{(x-0)^2+(y-0)^2}\\
&=\sqrt{x^2+y^2}
\end{align*}$$
Since $P$ is a point on the graph of $y=\frac{1}{x}$ , substitute $\frac{1}{x}$ for $y$.
Then,
$$d=\sqrt{x^2+\left(\frac{1}{x}\right)^2}=\sqrt{x^2+\frac{1}{x^2}}$$
Therefore, the distance $d$ from point $P$ to the origin expressed as a function of $x$ is:
$$d(x)=\sqrt{x^2+\frac{1}{x^2}}$$
(b) Use a graphing utility to graph $d(x)$. Refer to the graph below.
(c) From the graph it is clear that when the value of $x$ is $\pm 1$, $d$ is smallest.
