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What is the smallest distance between the origin and a point on the graph of y = 1/2*(x^2 - 8).

 Mar 2, 2024
 #1
avatar+129690 
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Let the point we want  be    (x , (1/2)(x^2 - 8))

 

Using the square of the distance

 

D^2  =  x^2  +   [ (1/2)(x^2 -8) ]^2

 

Take the derivative and set to 0

 

2x + (x^2 - 8)(2x)  = 0

 

2x + 2x^3 - 16x  = 0

 

2x^3 - 14x  =  0

 

x^3 - 7x   = 0

 

x( x^2 - 7)  = 0

 

x = 0   (reject)

 

x^2 - 7  = 0

 

x = sqrt (7)

 

The point  is  ( sqrt 7 , -1/2)

 

The distance  is  sqrt  [ (sqrt 7)^2  + (-1/2)^2 ] =  sqrt [ 7 + 1/4 ] =  sqrt (29 ) / 2

 

cool cool cool

 Mar 3, 2024

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