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

#1
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If we graph it on desmos, just to get a visual, We can see that the x intercepts are the closest point. Plugging in y = 0,

$$0=\frac{1}{\sqrt{2}}\left(x^{2}-3\right) \\ \frac{x^2}{\sqrt{2}} - \frac{3}{\sqrt{2}} = 0 \\ x^2 - 3 = 0 \\ x^2 = 3 \\ x = \pm\sqrt{3}$$

So the smallest distance is $\sqrt{3}$

Jul 8, 2021
#2
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you can't always jump to conclusions like that just by looking at the graph. :)

anyway

the distance from the graph to the origin can be represented like this (by the Pythagorean theorem):

$$\sqrt{x^2+(\frac{1}{\sqrt{2}}(x^2-3))^2}\\=\sqrt{x^2+\frac{(x^2-3)^2}{2}}\\= \sqrt{x^2+\frac{x^4-6x^2+9}{2}}\\=\sqrt{\frac{x^4-4x^2+9}{2}}\\ =\sqrt{\frac{x^4-4x^2+4+9-4}{2}}\\=\sqrt{\frac{(x^2-2)^2+5}{2}}$$

Since $$(x^2-2)^2$$ will always be nonnegative for real numbers, the best you can do to minimize it is to set it equal to 0, and you will obtain your final answer:

$$\sqrt{\frac{0+5}{2}}=\boxed{\sqrt{\frac{5}{2}}}$$

for completeness, if you want to know the x value for which this minimum distance occurs, you can just solve an equation:

$$(x^2-2)^2=0\\x^2-2=0\\x=\pm\sqrt{2}$$

textot  Jul 8, 2021
#3
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oh oops wow i mess that up lol

nice solution!!

Awesomeguy  Jul 8, 2021