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:
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