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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+128474 \(y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right)\)
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
