What point on the graph y = \sqrt{x} is closest to (4,0).
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If that gives you head ache you probably always have head ache, ha ha.
Just kidding.
(x , sqrt(x)) are the points on the graph
(4, 0) is the given point
the distance between those points, squared is
(x-4)² + (sqrt(x))² = (x-4)² + |x|, and we have to minimize that
for positive x this is (x-4)² + x = x² - 7x + 16
the minimum is when we derive and put the derivative zero :
2x - 7 = 0 => x=3.5
for negative x this is (x-4)² - x = x² - 9x + 16
=> 2x - 9 = 0 => x = 4.5, but this is not negative,
so we have
(3.5, sqrt(3.5)) as nearest point
Just kidding.
(x , sqrt(x)) are the points on the graph
(4, 0) is the given point
the distance between those points, squared is
(x-4)² + (sqrt(x))² = (x-4)² + |x|, and we have to minimize that
for positive x this is (x-4)² + x = x² - 7x + 16
the minimum is when we derive and put the derivative zero :
2x - 7 = 0 => x=3.5
for negative x this is (x-4)² - x = x² - 9x + 16
=> 2x - 9 = 0 => x = 4.5, but this is not negative,
so we have
(3.5, sqrt(3.5)) as nearest point
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D^2 = (4 - x)^2 + (0 - sqrt(x))^2
Find when dD/dx = 0
D^2 = (4 - x)^2 + (-sqrt(x))^2
D^2 = 16 - 8x + x^2 + x
D^2 = x^2 - 7x + 16
2D * dD/dx = 2x - 7
dD/dx = 0
2D * 0 = 2x - 7
0 = 2x - 7
2x = 7
x = 7/2
y = sqrt(7/2)
(7/2 , sqrt(7/2))
Find when dD/dx = 0
D^2 = (4 - x)^2 + (-sqrt(x))^2
D^2 = 16 - 8x + x^2 + x
D^2 = x^2 - 7x + 16
2D * dD/dx = 2x - 7
dD/dx = 0
2D * 0 = 2x - 7
0 = 2x - 7
2x = 7
x = 7/2
y = sqrt(7/2)
(7/2 , sqrt(7/2))