Find the point P on the graph of the function y=\sqrt{x} closest to the point (2,0)

at x, the tangent has slope 1/(2√x), so the normal has slope -2√x

So, the normal line at (k,√k) is

y-√k = -2√k (x-k)
Since (2,0) is on that line,

0-√k = -2√k(2-k)
1/2 = 2-k
k=3/2

The point is (3/2,√(3/2))
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or, just using the distance, we want the distance

d = √((2-x)^2+(0-√x)^2)
= √(x^2-3x+4)
to be a minimum

dd/dx = (2x-3) / 2√(x^2-3x+4)
dd/dx=0 at x = 3/2
The desired point is (3/2,√(3/2))

To find the point P on the graph of the function y = √x closest to the point (2,0), we can use the distance formula between two points.

Let P = (x, √x) be the point on the graph of the function y = √x. The distance between P and (2,0) is given by the formula:

d = √((x - 2)^2 + (√x - 0)^2)

To find the minimum distance, we can take the derivative of the distance function and set it equal to zero:

d' = 0

Differentiating the distance formula:

d' = (x - 2) / √((x - 2)^2 + (√x)^2) + (1/2) * (2√x) / √((x - 2)^2 + (√x)^2) - (1/2) * 2(x - 2) / √((x - 2)^2 + (√x)^2)

Setting d' equal to zero:

(x - 2) / √((x - 2)^2 + (√x)^2) + √x / √((x - 2)^2 + (√x)^2) - (x - 2) / √((x - 2)^2 + (√x)^2) = 0

Cancelling out the denominators:

(x - 2) + √x - (x - 2) = 0

Simplifying:

√x = 2

Squaring both sides:

x = 4

So, the x-coordinate of the point P is 4. To find the y-coordinate, we substitute x = 4 into the equation y = √x:

y = √4 = 2

Therefore, the point P on the graph of the function y = √x closest to the point (2,0) is P(4, 2).

To find the point P on the graph of the function y = √x closest to the point (2,0), we can use the distance formula between two points. The formula for finding the distance between two points (x1, y1) and (x2, y2) is:

d = √[(x2 - x1)^2 + (y2 - y1)^2]

In this case, the point (2,0) is our reference point (x1, y1), and the point P on the graph of y = √x is our target point (x2, y2).

Let's denote the x-coordinate of point P as xP, and the y-coordinate as yP (which is equal to √xP). We need to find the values of xP and yP that minimize the distance d.

Using the distance formula, we have:

d = √[(xP - 2)^2 + (√xP - 0)^2]

To minimize the distance, we can minimize the expression inside the square root. So, let's minimize:

(xP - 2)^2 + (√xP - 0)^2

Expanding the expression, we get:

xP^2 - 4xP + 4 + xP

Combining like terms, we have:

xP^2 - 3xP + 4

To find the minimum value of this quadratic function, we can use calculus. Taking the derivative of the function with respect to xP and setting it equal to zero will give us the critical point(s).

d/dxP (xP^2 - 3xP + 4) = 0

2xP - 3 = 0

Solving for xP, we find:

xP = 3/2

Now that we have the x-coordinate of point P, we can substitute it back into the equation yP = √xP to find the y-coordinate:

yP = √(3/2) = √(3/2) * √(2/2) = √(3/4) = √3/√4 = √3/2

Therefore, the point P on the graph of y = √x closest to the point (2,0) is approximately (3/2, √3/2).