What is the smallest distance between the origin and a point on the graph below.
I can see that the graph is a concave up parabola'
So I know a smallest distance exists, there will be no maximum distance.
\(y=\dfrac{1}{\sqrt{2}}\left(x^2-8\right)\\ \left ( x,\dfrac{1}{\sqrt{2}}\left(x^2-8\right)\right) \text{ is a general point on this graph}\\ \text{I need the distance of this point from (0,0)}\\ d^2=x^2+[\frac{1}{\sqrt{2}}\left(x^2-8\right)]^2\\ d^2=x^2+\frac{1}{2}\left(x^2-8\right)^2\\ \)
d will be minimum when d^2 is minimum. So to avoid confusion I will let v=d^2
\(v=x^2+\frac{1}{2}\left(x^2-8\right)^2\\\)
\(\frac{dv}{dx }=2x+(x^2-8)*2x\\ \frac{dv}{dx }=2x(1+x^2-8)\\ \frac{dv}{dx }=2x(x^2-7)\\ \frac{dv}{dx }=0\;\;when \;\;x=0\;\;or\;\;x=\pm\sqrt7\\ \)
When x=0 v=32
When x= +/-sqrt7 v=7+0.5=7.5
Clearly the min is when v=7.5
So the minimum distance is \(\sqrt{7.5}\)
LaTex
y=\dfrac{1}{\sqrt{2}}\left(x^2-8\right)\\
\left ( x,\dfrac{1}{\sqrt{2}}\left(x^2-8\right)\right) \text{ is a general point on this graph}\\
\text{I need the distance of this point from (0,0)}\\
d^2=x^2+[\frac{1}{\sqrt{2}}\left(x^2-8\right)]^2\\
d^2=x^2+\frac{1}{2}\left(x^2-8\right)^2\\
v=x^2+\frac{1}{2}\left(x^2-8\right)^2\\
