3) which point on the graph of y = √x is closest to the point (4, 0)?

Explanation:

Step1: Define the distance formula

Let a point on the curve $y = \sqrt{x}$ be $(x,y)=(x,\sqrt{x})$. The distance $d$ between $(x,\sqrt{x})$ and $(4,0)$ is given by the distance formula $d=\sqrt{(x - 4)^2+(\sqrt{x}-0)^2}=\sqrt{(x - 4)^2+x}$. To simplify the calculations, we can minimize the square of the distance function $D(x)=(x - 4)^2+x$.

Step2: Expand the function

Expand $D(x)$: [ \begin{align*} D(x)&=x^{2}-8x + 16+x\ &=x^{2}-7x + 16 \end{align*} ]

Step3: Find the derivative

Differentiate $D(x)$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $D^\prime(x)=2x-7$.

Step4: Set the derivative equal to zero

Set $D^\prime(x) = 0$ to find the critical points: [ \begin{align*} 2x-7&=0\ 2x&=7\ x&=\frac{7}{2} \end{align*} ]

Step5: Find the corresponding y - value

Since $y=\sqrt{x}$, when $x = \frac{7}{2}$, $y=\sqrt{\frac{7}{2}}$.

Answer:

$(\frac{7}{2},\sqrt{\frac{7}{2}})$