To determine the point on the graph of $y = \sqrt{x}$ that is closest to the point $(7, 0)$, we can approach the problem by minimizing the distance between a general point on the curve $(x, \sqrt{x})$ and the point $(7, 0)$.

Solution Steps:

1. Define the Distance Function: The distance $d$ between any point $(x, \sqrt{x})$ on the curve and $(7, 0)$ is given by: $d = \sqrt{(x - 7)^2 + (\sqrt{x})^2}$ Simplifying this, we get: $d = \sqrt{(x - 7)^2 + x}$

2. Minimize the Distance: Since the square root is a monotonically increasing function, minimizing $d$ is equivalent to minimizing $d^2$, so we define: $f(x) = (x - 7)^2 + x$ We now need to find the minimum of $f(x)$ for $x \geq 0$.

3. Differentiate and Set to Zero: Differentiate $f(x)$ with respect to $x$ to find the critical points. $f'(x) = 2(x - 7) + 1 = 2x - 14 + 1 = 2x - 13$ Set $f'(x) = 0$: $2x - 13 = 0 \implies x = 6.5$

4. Check the Point on the Graph: For $x = 6.5$, $y = \sqrt{6.5}$.

   Therefore, the point on the curve closest to $(7, 0)$ is $(6.5, \sqrt{6.5})$.

5. Verify the Minimum: Since $f(x)$ is a quadratic function that opens upwards, this critical point at $x = 6.5$ provides the minimum distance.

Answer:

The point on the graph of $y = \sqrt{x}$ that is closest to $(7, 0)$ is $(6.5, \sqrt{6.5})$.

Would you like more details on any part of this solution, or do you have any questions?

Here are five additional questions to consider:

1. How would the process change if the curve were different, like $y = x^2$?
2. What if the point were not $(7, 0)$ but another arbitrary point $(a, b)$?
3. Can this method be generalized to find the closest point on any function to a given point?
4. How does minimizing $d^2$ simplify the calculations compared to minimizing $d$?
5. What is the significance of taking the derivative and setting it to zero in optimization problems?

Tip: For optimization problems involving distance, minimizing the square of the distance function can simplify calculations without affecting the result, since distance and squared distance reach their minimums at the same point.