Which point on the curve y=square root x-2+1 is closest to the
point (4, 1)? What is this minimum distance?

To find the point on the curve \( y = \sqrt{x - 2} + 1 \) that is closest to the point \( (4, 1) \), we need to minimize the distance between any point \( (x, y) \) on the curve and the given point \( (4, 1) \).

### Step 1: Distance Formula

The distance \( D \) between a point \( (x, y) \) and \( (4, 1) \) is given by:

\[
D = \sqrt{(x - 4)^2 + (y - 1)^2}
\]

Since \( y = \sqrt{x - 2} + 1 \), we substitute this into the distance formula:

\[
D(x) = \sqrt{(x - 4)^2 + (\sqrt{x - 2} + 1 - 1)^2}
\]

Simplify the expression:

\[
D(x) = \sqrt{(x - 4)^2 + (\sqrt{x - 2})^2}
\]

\[
D(x) = \sqrt{(x - 4)^2 + (x - 2)}
\]

\[
D(x) = \sqrt{x^2 - 8x + 16 + x - 2}
\]

\[
D(x) = \sqrt{x^2 - 7x + 14}
\]

### Step 2: Minimizing the Distance

To find the minimum distance, we need to minimize \( D(x) \). However, since the square root is a monotonically increasing function, we can instead minimize the squared distance \( D^2(x) = x^2 - 7x + 14 \).

Let:

\[
f(x) = x^2 - 7x + 14
\]

### Step 3: Find the Critical Points

To find the critical points, we take the derivative of \( f(x) \) with respect to \( x \) and set it to zero:

\[
f'(x) = 2x - 7
\]

Set \( f'(x) = 0 \):

\[
2x - 7 = 0
\]

\[
x = \frac{7}{2} = 3.5
\]

### Step 4: Verify if \( x = 3.5 \) is within the domain

The curve \( y = \sqrt{x - 2} + 1 \) is defined for \( x - 2 \geq 0 \) or \( x \geq 2 \). Since \( 3.5 \geq 2 \), it is within the domain.

### Step 5: Find the Corresponding \( y \) and the Minimum Distance

Substitute \( x = 3.5 \) into the curve equation to find \( y \):

\[
y = \sqrt{3.5 - 2} + 1 = \sqrt{1.5} + 1
\]

Thus, the point on the curve closest to \( (4, 1) \) is \( \left(3.5, \sqrt{1.5} + 1\right) \).

Now, calculate the minimum distance:

\[
D_{\text{min}} = \sqrt{(3.5 - 4)^2 + (\sqrt{1.5} + 1 - 1)^2}
\]

\[
D_{\text{min}} = \sqrt{(-0.5)^2 + (\sqrt{1.5})^2}
\]

\[
D_{\text{min}} = \sqrt{0.25 + 1.5}
\]

\[
D_{\text{min}} = \sqrt{1.75}
\]

\[
D_{\text{min}} \approx 1.32
\]

### Conclusion

The point on the curve \( y = \sqrt{x - 2} + 1 \) closest to \( (4, 1) \) is approximately \( \left(3.5, \sqrt{1.5} + 1\right) \). The minimum distance between the point \( (4, 1) \) and the curve is approximately **1.32** units.

Would you like a detailed breakdown of any part, or have any questions?

### Related Questions

1. How do you find the distance between two points in a coordinate plane?
2. What is the domain of the function \( y = \sqrt{x - 2} + 1 \)?
3. How can the derivative test determine the minimum or maximum of a function?
4. How do you calculate the square root of a non-perfect square manually?
5. What are other methods for minimizing functions in calculus?

### Tip: 
Always check the domain of the functions involved in problems to avoid undefined values and incorrect results.

Learn how to find the point on the curve y = sqrt(x - 2) + 1 that is closest to the point (4, 1) using calculus and the distance formula. Discover the steps involved in minimizing the distance function and calculating the minimum distance. Suitable for advanced high school students studying calculus and functions.

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