Answer
(a) $d(x)=\sqrt{x^2-x+1}$
(b) Refer to the graph below.
(c) $d$ is smallest when $x=0.5$.
Work Step by Step
(a) The distance $d$ from $P(x,y)$ to the point $(1,0)$ is:
$$\begin{align*}
d&=\sqrt{(x-x_1)^2+(y-y_1)^2}\\
&=\sqrt{(x-1)^2+(y-0)^2}\\
&=\sqrt{x^2-2x+1+y^2}
\end{align*}$$
Since $P$ is a point on the graph of $y=\sqrt{x}$ , substitute $\sqrt{x}$ for $y$.
Then,
$$d=\sqrt{x^2-2x+1+(\sqrt{x})^2}=\sqrt{x^2-2x+1+x}=\sqrt{x^2-x+1}$$
Therefore, the doistance of the point $P$ from $(1, 0)$ expressed as a function of $x$ is:
$$d(x)=\sqrt{x^2-x+1}$$
(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 $0.5$ , $d$ is smallest.
