Question

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Answer 1

The closest point is (3.5, 1.9) and the distance is 1.96 units

How to determine the point and the distance?

The coordinate is given as:

(4, 0)

The equation of the function is

y = √x

The distance between two points is calculated using

d = √(x2 - x1)^2 + (y2 - y1)^2

So, we have the following points

(x1, y1) = (4, 0) and (x2, y2) = (x, 0)

This gives

d = √(x - 4)^2 + (√x - 0)^2

Evaluate the difference

d = √(x - 4)^2 + (√x)^2

Evaluate the exponent

d = √x^2 - 8x + 16 + x

Evaluate the like terms

d = √x^2 - 7x + 16

Next, we differentiate using a graphing calculator

d' = (2x - 7)/[2√(x^2 - 7x + 16)]

Set to 0

(2x - 7)/[2√(x^2 - 7x + 16)] = 0

Cross multiply

2x - 7 = 0

Add 7 to both sides

2x = 7

Divide by 2

x = 3.5

So, we have:

Substitute x = 3.5 in y = √x

y = √3.5

Evaluate

y = 1.9

So, the point is (3.5, 1.9)

The distance is then calculated as:

d = √(x2 - x1)^2 + (y2 - y1)^2

This gives

d = √(3.5 - 4)^2 + (1.9 - 0)^2

Evaluate

d = 1.96

Hence, the closest point is (3.5, 1.9) and the distance is 1.96 units

Read more about distance at:

brainly.com/question/23848540

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