What is the smallest distance between the origin and a point on the graph of $y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right)?$

Edit:

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Guest Jan 12, 2018

#1**+1 **

\(y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right)\)

The distance is given by :

√ [ x^2 + (x^4 - 6x^2 + 9) / 2] =

√ [ 2x^2 + x^4 - 6x^2 + 9] * √[ 1/2] =

√ [ x^4 - 4x^2 + 9] * √[1/2 ] =

[ x^4/2 - 2x^2 + 9/2 ] ^(1/2)

Taking the derivative of this and setting to 0 we have

[ 2x^3 - 4x] / [ 2 [ x^4/2 - 2x^2 + 9/2 ] ^(1/2)] = 0

Multiply through by the denominator and we have that

[ 2x^3 - 4x] = 0 factor

2x [ x^2 - 2] = 0

Solving this for x gives x = 0, x = √2 and x = -√ 2

We could take the second derivative and determine which of these values gives a minimum, but it's messy.....a better approach is to graph the given function and the circle x^2 + y^2 = 9/2 ....the reason for this is that when x = 0 , the associated function point lies on the circle

See the graph here : https://www.desmos.com/calculator/upxjjsafyd

But....when x = ± √ 2.....the associated points on the function (±√ 2, - 1/ √ 2) will fall inside this circle

So....using either point, the minimum distance is

√ [ 2 + 1/2 ] =

√ [ 2.5 ] units ≈ 1.58 units

CPhill Jan 12, 2018